Coding Interview Patterns Flashcards: Signals, Invariants & Complexity
Practice coding interview patterns with 210 English flashcards on problem signals, invariants, complexity, edge cases, and choosing between approaches. Language-agnostic DSA review to pair with hands-on coding.
Про цю колоду
Coding interview patterns become useful when you can explain why an approach fits. These 210 English flashcards practice recognizing problem signals, stating invariants, checking prerequisites, comparing approaches, and diagnosing a broken assumption.
The sequence starts with constraints and complexity, then moves through arrays, strings, hashing, two pointers, sliding windows, prefix sums, binary search, intervals, stacks, linked lists, trees, heaps, tries, graphs, union-find, topological sorting, backtracking, greedy choices, dynamic programming, and bit techniques. Each topic builds from its basic idea toward the conditions that make it work.
Cards map short scenarios to candidate patterns, named patterns to invariants and costs, and failed assumptions to explanations or alternatives. Selected contrast prompts distinguish neighboring ideas such as subarrays and subsequences, BFS and DFS, or greedy choice and dynamic programming. Mechanical reverse cards, problem-number recall, language-specific API memorization, and full solution listings are excluded because they add little to this reasoning-focused review. Answer each prompt before turning the card, then test the idea by coding a fresh example.
This is a language-agnostic companion to practical DSA exercises. It complements the Blind 75 Python solutions deck by teaching reusable reasoning across problems. It was authored independently in English, not translated or paraphrased from another catalog package.
The questions, answers, organization, metadata, and original generated cover are released under CC0 1.0 to the extent applicable rights exist. Algorithm facts are common knowledge; no commercial deck text or third-party media was copied. This is an independent study resource with no affiliation to an employer or interview platform.
Картки в цій колоді
Картка 1
Питання
Which input constraints should you clarify before choosing an interview algorithm?
Відповідь
Input size, value range, ordering, duplicates, allowed mutations, and the exact output. These determine which operations are affordable and which assumptions are valid.
Картка 2
Питання
What does a loop invariant describe?
Відповідь
A property that holds at a defined point in every iteration. Show it holds initially, survives an iteration, and implies the result when the loop ends.
Картка 3
Питання
When analyzing nested loops, why can multiplying their written bounds overestimate runtime?
Відповідь
The loops may share progress. If an inner pointer only advances across the input once, its total work can be O(n), even inside an outer loop.
Картка 4
Питання
What is the difference between auxiliary space and total space?
Відповідь
Auxiliary space counts extra working storage. Total space also includes the input and, depending on the stated convention, the output. State which measure you report.
Картка 5
Питання
What does amortized O(1) mean for an operation sequence?
Відповідь
The total cost of m operations is O(m), even if some individual operations are expensive. It is a sequence-wide bound, not a probability claim.
Картка 6
Питання
How does a counterexample help evaluate a proposed greedy rule?
Відповідь
One valid input where the rule fails disproves it. Try small cases that force a locally attractive choice to block a better later choice.
Картка 7
Питання
Why should an algorithm's correctness argument address termination separately?
Відповідь
Preserving the right property is not enough if the loop never stops. Identify a bounded measure that moves strictly toward termination.
Картка 8
Питання
What runtime lower bound follows from returning k separate results?
Відповідь
At least Ω(k) time to emit them, or more if each result has multiple elements. Include output size when analyzing enumeration algorithms.
Картка 9
Питання
What should you say when quoting expected O(1) hash-table lookup?
Відповідь
It assumes a suitable hash distribution and controlled load factor. It is not a worst-case guarantee; hashing a long key can also cost time.
Картка 10
Питання
Which edge cases best expose index and boundary errors?
Відповідь
Empty input, one element, all equal elements, and answers at either end. Also check the smallest input that enters each branch.
Картка 11
Питання
Why can sorting be an invalid optimization even when it reduces later search work?
Відповідь
Sorting may destroy required order or original indices. Preserve the needed information or choose an approach that respects the input contract.
Картка 12
Питання
What does an exchange argument establish in a greedy proof?
Відповідь
That an optimal solution can be changed to include the greedy choice without making its objective worse. The remaining problem must still fit the same reasoning.
Картка 13
Питання
Why can recursion use O(n) space even without an explicit collection?
Відповідь
Each active call occupies a stack frame. A chain of n calls can therefore need O(n) auxiliary space.
Картка 14
Питання
What evidence should accompany a faster solution after presenting brute force?
Відповідь
Name the repeated work or discarded search space, explain why the shortcut is safe, and give the resulting time and space bounds.
Картка 15
Питання
What makes an array useful when a problem repeatedly accesses positions by index?
Відповідь
Constant-time indexed access in the usual RAM model. Inserting or removing near the front can still require shifting O(n) elements.
Картка 16
Питання
What should 'one character' mean before solving a string problem?
Відповідь
Clarify whether the unit is a byte, code unit, Unicode code point, or user-perceived character. Indexing and length depend on that choice.
Картка 17
Питання
An unsorted array needs a duplicate-existence check. Which structure fits?
Відповідь
A hash set of values seen so far. Stop when a value is already present; expected O(n) time and O(n) space.
Картка 18
Питання
For two-sum on an unsorted array, what should a hash map store?
Відповідь
Previously seen values mapped to their indices. For each value x, look for target − x before inserting x, so the same position is not reused.
Картка 19
Питання
When does a frequency array beat a hash map for counting?
Відповідь
When keys come from a small known integer or character range. Direct indexing gives predictable access with space proportional to that range.
Картка 20
Питання
Why is repeated concatenation risky when constructing a long immutable string?
Відповідь
Each append may copy the existing prefix, producing quadratic total work. Collect pieces and join them, or use an appropriate mutable builder.
Картка 21
Питання
How can you group anagrams without comparing every pair of words?
Відповідь
Map a canonical character signature to a group. Sorted characters work; a frequency tuple works when the alphabet and character rules are fixed.
Картка 22
Питання
What information does a set lose compared with a frequency map?
Відповідь
Multiplicity. A set can answer whether a value exists but cannot distinguish one occurrence from several.
Картка 23
Питання
How can a hash set support finding the longest consecutive integer run in expected O(n) time?
Відповідь
Start scanning a run only at values whose predecessor is absent. Each distinct value then belongs to one forward scan; iterate distinct values.
Картка 24
Питання
Why can a mutable object be a dangerous hash-map key?
Відповідь
Changing fields used by its hash or equality can make the stored entry unreachable by ordinary lookup. Use immutable keys or stable key values.
Картка 25
Питання
An array contains only integers from 0 through k. When is counting sort attractive?
Відповідь
When k is small enough: count each value and reconstruct the order in O(n + k) time with O(k) count storage. General arbitrary keys need another approach.
Картка 26
Питання
How can a single scan find both the minimum value and its earliest index?
Відповідь
Keep the current minimum and index; update only on a strictly smaller value. Updating on equality would select a later occurrence.
Картка 27
Питання
How do you compare two strings as multisets of characters?
Відповідь
Compare character counts under the same character interpretation. Order does not matter, but every character's multiplicity does.
Картка 28
Питання
Why does a hash collision not imply that two keys are equal?
Відповідь
A hash compresses many possible keys into fewer codes. A correct table also checks key equality when resolving collisions.
Картка 29
Питання
When is sorting a useful preprocessing step for detecting duplicate values?
Відповідь
When reordering is allowed and O(n log n) time is acceptable. Equal values become adjacent, so a scan finds duplicates without a separate hash set.
Картка 30
Питання
What is the key distinction between a subarray and a subsequence?
Відповідь
A subarray is contiguous. A subsequence preserves relative order but may skip positions. Sliding-window methods usually depend on contiguity.
Картка 31
Питання
For an unsorted two-sum query, how do hashing and sorting trade off?
Відповідь
Hashing gives expected O(n) time with O(n) extra storage. Sorting plus two pointers costs O(n log n) time and requires care with original indices and mutation.
Картка 32
Питання
How can a frequency map detect whether any permutation of a string can be a palindrome?
Відповідь
Count odd frequencies. At most one character may have an odd count; all other occurrences must form mirrored pairs.
Картка 33
Питання
Why must compound hash keys encode boundaries unambiguously?
Відповідь
Naive concatenation can merge different tuples into the same key, such as (1, 23) and (12, 3). Use tuples or a length-aware encoding.
Картка 34
Питання
What does coordinate compression preserve about numeric values?
Відповідь
Their relative order and equality, by replacing distinct sorted values with ranks. It does not preserve numeric distances or sums.
Картка 35
Питання
How can you compute products except self without division?
Відповідь
Combine the product strictly before each position with the product strictly after it. Prefix and suffix passes handle zeros; use a numeric type large enough for the products.
Картка 36
Питання
Why can a count of matching pairs overflow even when every input value fits in an integer?
Відповідь
The number of matching pairs can grow quadratically with the input length. Size the result type for the count, not just the input values.
Картка 37
Питання
A sorted array needs a pair with a target sum. Which search pattern fits?
Відповідь
Opposite-end two pointers. Compare the endpoint sum with the target and move the endpoint that can move the sum in the needed direction.
Картка 38
Питання
What invariant supports in-place removal of unwanted array values with read and write pointers?
Відповідь
The prefix before write contains exactly the retained values from the processed input, in order. Read scans every input position once.
Картка 39
Питання
How can two pointers check a palindrome without constructing a reversed string?
Відповідь
Compare matching elements from opposite ends and move inward. Any mismatch rejects it; the pointers meeting or crossing completes the check.
Картка 40
Питання
When does a fixed-size sliding window apply?
Відповідь
When every candidate is a contiguous block of the same length and its summary can be updated as one element enters and one leaves.
Картка 41
Питання
What invariant should a longest-window algorithm restore after adding a new rightmost element?
Відповідь
The current window satisfies the required constraint. Move the left boundary and update state until validity returns, then consider its length.
Картка 42
Питання
Why is moving the left pointer safe when a sorted-array endpoint sum is too small?
Відповідь
With that left value, every candidate at or before the current right endpoint gives an equally small or smaller sum. The left position cannot form a valid pair.
Картка 43
Питання
How does a three-way partition maintain separate regions?
Відповідь
Track a low region, a middle region, an unclassified region, and a high region. Each step shrinks the unclassified region by placing one element correctly.
Картка 44
Питання
How do you update the sum when a fixed-size window moves one position?
Відповідь
Subtract the outgoing value and add the incoming value. After initializing the first window, all moves together take O(n) time.
Картка 45
Питання
Why can a variable sliding window run in O(n) despite a nested shrink loop?
Відповідь
Each boundary advances at most n times. With constant-time state updates, the total number of additions and removals is linear.
Картка 46
Питання
For the longest substring without repeated characters, what window state is useful?
Відповідь
Character counts or last-seen positions. Move the left boundary past the conflicting occurrence while ensuring it never moves backward.
Картка 47
Питання
Why does opposite-end target-sum search fail on a generally unsorted array?
Відповідь
Moving an endpoint no longer changes the sum predictably. The algorithm can discard a valid pair because the order-based elimination proof is missing.
Картка 48
Питання
A positive-number array needs the shortest nonempty subarray with sum at least a positive T. When should the left boundary move?
Відповідь
While the current sum is at least T, record the length and remove the leftmost value. Positive values make further shrinking reduce the sum predictably.
Картка 49
Питання
For windows with at most k distinct values, what must happen when an outgoing count becomes zero?
Відповідь
Remove that value from the active distinct set or decrement the distinct counter. A zero count must not still count as present.
Картка 50
Питання
What changes when a fixed-window length exceeds the input length?
Відповідь
There is no complete window. Return the contract's empty or missing-result value instead of treating a partial block as a valid candidate.
Картка 51
Питання
How do you merge two sorted arrays with forward pointers?
Відповідь
Repeatedly take the smaller unconsumed element, then append the remaining suffix. Each element is consumed once, giving O(m + n) time.
Картка 52
Питання
Why must a minimum-cover substring track multiplicities of required characters?
Відповідь
A target can require the same character more than once. A set of required characters would mark an undersupplied window as complete.
Картка 53
Питання
How can counting subarrays with exactly k distinct values, for k ≥ 1, use an at-most helper?
Відповідь
Compute atMost(k) − atMost(k − 1). The helper counts all valid subarrays ending at each right boundary after restoring the distinct-value limit.
Картка 54
Питання
Why can a negative value break the usual shortest-sum sliding-window argument?
Відповідь
Removing it increases the sum, and adding one can decrease the sum. The monotonic relation between window size and sum no longer holds.
Картка 55
Питання
What does 'fast and slow pointers' mean when removing duplicates from a sorted array?
Відповідь
A read pointer scans candidates while a write pointer marks the next unique slot. Sorting makes equal values adjacent, so the retained prefix can stay compact.
Картка 56
Питання
What invariant prevents overwriting unread data during a backward merge into spare array capacity?
Відповідь
The suffix after the write pointer already contains the largest merged elements. Filling from the end leaves the remaining source elements unread and intact.
Картка 57
Питання
Why is 'find a contiguous range' alone insufficient to justify a variable sliding window?
Відповідь
You also need a safe boundary-movement rule. Check how adding and removing elements affect the condition; contiguity by itself gives no such guarantee.
Картка 58
Питання
After restoring an at-most window's validity, why are there right − left + 1 valid subarrays ending at right?
Відповідь
Every suffix starting between left and right is valid when removing elements cannot violate the constraint. Count those starts, including the one-element suffix.
Картка 59
Питання
How can you avoid duplicate value pairs in a sorted two-pointer enumeration?
Відповідь
After emitting a pair, skip equal values on both sides. First confirm the output wants unique value pairs, since index-pair counting needs different handling.
Картка 60
Питання
When does a character-frequency sliding window detect an anagram of a pattern?
Відповідь
When the window has the pattern's length and identical character counts. Maintain count differences or a mismatch counter as the window moves.
Картка 61
Питання
What does a prefix-sum array P mean when P[0] = 0?
Відповідь
P[i] is the sum of the first i input elements. The extra zero represents the empty prefix and makes ranges beginning at index 0 work uniformly.
Картка 62
Питання
When is binary search valid on a Boolean predicate over ordered candidates?
Відповідь
When the predicate changes at most once, such as false then true. The search uses that monotonic boundary to discard a whole interval.
Картка 63
Питання
Which workload favors a difference array?
Відповідь
Many range additions followed by final value reconstruction. Mark each range's start and end changes, then take one prefix sum.
Картка 64
Питання
In a lower-bound search over [lo, hi), what does hi initially equal for an n-element array?
Відповідь
n, an exclusive boundary. The result can equal n when no element is at least the target, so do not index the array without checking.
Картка 65
Питання
For a static array, how do prefix sums answer the half-open range [l, r)?
Відповідь
Return P[r] − P[l]. Building P takes O(n) time and space; each range-sum query then takes O(1).
Картка 66
Питання
How can prefix sums count subarrays whose sum equals k when negative values are allowed?
Відповідь
For each current prefix p, add the number of earlier prefixes equal to p − k, then record p. A frequency map gives expected O(n) time.
Картка 67
Питання
Why is binary-searching an answer different from binary-searching an input array?
Відповідь
The candidates are possible result values. A feasibility check tells which side contains the boundary, even if the original input is unsorted.
Картка 68
Питання
How does a difference array encode an addition of v to [l, r)?
Відповідь
Add v at l and subtract v at r, using a boundary slot when needed. The reconstructed prefix totals apply v only within that range.
Картка 69
Питання
For lower bound, how should equality with the target move the search boundary?
Відповідь
Move hi to mid. An equal element is a candidate, but an earlier equal or qualifying element may still exist.
Картка 70
Питання
Why initialize the prefix-frequency map with zero appearing once?
Відповідь
It represents the empty prefix before the array. This lets a subarray starting at index 0 contribute to the count.
Картка 71
Питання
Why are plain prefix sums inconvenient for many interleaved point updates and range-sum queries?
Відповідь
Changing one value can invalidate a long suffix of prefix sums. A Fenwick tree or segment tree can support both operations in O(log n).
Картка 72
Питання
How can you binary-search the minimum capacity needed to finish ordered work within a deadline?
Відповідь
Define whether a capacity suffices, prove larger capacities remain feasible, bracket a feasible answer, and search for the first feasible capacity.
Картка 73
Питання
What must every binary-search iteration do to guarantee termination?
Відповідь
Strictly shrink the candidate interval while preserving the boundary invariant. Mixing inclusive and exclusive update rules can leave the same interval unchanged.
Картка 74
Питання
For the longest subarray with a specified sum, which occurrence of each prefix sum should you retain?
Відповідь
The earliest index. For a later endpoint, it gives the longest matching span; overwriting it with a later occurrence can shorten the answer.
Картка 75
Питання
What runtime should you report for binary search with a nonconstant feasibility check?
Відповідь
O(C log R), where C is one check's cost and R is the number of discrete candidates. Include any preprocessing separately.
Картка 76
Питання
How can prefix sums turn a longest balanced binary subarray into an equal-prefix problem?
Відповідь
Map one symbol to +1 and the other to −1. Equal prefix sums enclose a zero-sum range with equal counts of the two symbols.
Картка 77
Питання
What is upper bound in a sorted array?
Відповідь
The first position whose value is strictly greater than the target, or n if none exists. Lower bound instead finds the first value at least the target.
Картка 78
Питання
Why must a prefix-sum counting algorithm query before recording the current prefix?
Відповідь
Recording first can count the empty subarray ending at the current boundary, especially for target zero. Query only earlier prefixes for nonempty ranges.
Картка 79
Питання
How do you safely compute a midpoint in a fixed-width integer search?
Відповідь
Use lo + (hi − lo) / 2 with integer division when the nonnegative difference fits the type. Choose bounds or a wider type that also keep the subtraction safe.
Картка 80
Питання
Why can duplicate values degrade searching a rotated sorted array to O(n)?
Відповідь
Equal endpoints and midpoint can hide which side is sorted. Some cases permit discarding only one boundary element at a time.
Картка 81
Питання
What preprocessing usually simplifies merging overlapping intervals?
Відповідь
Sort by start coordinate. Keep the current merged interval and either extend its end or emit it when the next interval starts beyond it.
Картка 82
Питання
Which data structure matches nested bracket validation?
Відповідь
A stack of unmatched opening brackets. Each closing bracket must match the most recent unmatched opener, and the stack must be empty at the end.
Картка 83
Питання
A problem asks for each element's next greater element. Which pattern is promising?
Відповідь
A monotonic stack of unresolved positions. A new larger value resolves the smaller pending values it overtakes.
Картка 84
Питання
Why must interval endpoint conventions be explicit?
Відповідь
Touching endpoints overlap for closed intervals, but adjacent half-open intervals do not. The convention changes merge tests and event ordering.
Картка 85
Питання
How can a sweep line find the maximum number of simultaneous intervals?
Відповідь
Turn starts and ends into signed events, sort by coordinate, and track the running active count. Handle same-coordinate ties according to the endpoint convention.
Картка 86
Питання
Why is one pass after sorting enough to merge intervals?
Відповідь
No later interval starts earlier than the next one being inspected. Once that start is beyond the current end, future intervals cannot bridge the gap.
Картка 87
Питання
What information should a stack store for next-greater distances?
Відповідь
Indices, so the distance is currentIndex − previousIndex. Values alone do not identify positions or distinguish repeated occurrences.
Картка 88
Питання
Why is counting opening and closing brackets insufficient to validate their sequence?
Відповідь
Counts ignore order and nesting. A closing bracket may appear before its opener, or bracket types may cross despite balanced totals.
Картка 89
Питання
For half-open intervals [start, end), how should equal-time starts and ends affect room counts?
Відповідь
Process ends before starts, or aggregate their net change before evaluating the active count for the next segment. A room freed at time t can be reused at t.
Картка 90
Питання
Why is a monotonic-stack algorithm often O(n) even though one step can pop many items?
Відповідь
Each item is pushed once and popped at most once. Summed across the scan, stack operations are linear.
Картка 91
Питання
How can a stack help simplify an absolute filesystem path lexically?
Відповідь
Process components: ignore empty components and '.', pop for '..' when possible, and push ordinary names. This lexical result does not resolve symbolic links.
Картка 92
Питання
How can a monotonic deque find each sliding-window maximum?
Відповідь
Keep candidate indices in decreasing value order. Remove expired indices from the front and dominated values from the back; the front gives the maximum.
Картка 93
Питання
What mistake can lose coverage when merging an interval contained inside the current one?
Відповідь
Replacing the current end with the new end. Use the larger end so a nested interval cannot shrink the merged coverage.
Картка 94
Питання
For a strictly next-greater query, what should happen to an equal-valued stack entry?
Відповідь
Do not resolve it with the equal value. With a decreasing stack of unresolved indices, pop only when the new value is strictly greater.
Картка 95
Питання
What event allows a monotonic stack to finalize a rectangle in a histogram?
Відповідь
A shorter bar supplies a right limit for taller bars being popped. A popped bar's candidate span starts after the new stack top, or at index 0 if the stack is empty. Handle equal heights consistently.
Картка 96
Питання
What comparison detects overlap between two nonempty half-open intervals?
Відповідь
max(start1, start2) < min(end1, end2). A strict comparison excludes intervals that only touch.
Картка 97
Питання
How does a stack support evaluating a postfix arithmetic expression?
Відповідь
Push operands; for an operator, pop its right operand and then its left operand, compute, and push the result. Preserve order for subtraction and division.
Картка 98
Питання
Why can a newer value dominate an older value in a sliding-window maximum deque?
Відповідь
If the newer value is at least as large, it expires no earlier and is never worse as a future maximum. The older candidate can be removed.
Картка 99
Питання
How can you merge two already sorted lists of disjoint intervals to find their intersections?
Відповідь
Compare one interval from each list, emit any overlap, then advance the one with the earlier end. Total time is O(m + n).
Картка 100
Питання
Why must a histogram stack algorithm handle bars still pending after the scan?
Відповідь
Those bars may extend to the array's end and contain the largest rectangle. Flush them using the end boundary or a suitable sentinel.
Картка 101
Питання
What must you save before reversing a singly linked list node's next pointer?
Відповідь
Its original next node. Otherwise rewiring can lose access to the remaining list.
Картка 102
Питання
Why does a dummy head simplify linked-list insertion and deletion?
Відповідь
It supplies a predecessor even when the real head changes. The same pointer update can handle both the first node and interior nodes.
Картка 103
Питання
How do fast and slow pointers detect a cycle in a singly linked list?
Відповідь
Advance one pointer one step and the other two. A meeting implies a cycle; reaching null with the fast pointer means the list terminates.
Картка 104
Питання
What does a recursive binary-tree traversal use for auxiliary space?
Відповідь
O(h) call-stack space, where h is tree height. This is O(log n) for a balanced tree but O(n) for a chain.
Картка 105
Питання
During iterative list reversal, what do prev and current represent?
Відповідь
prev heads the reversed processed prefix; current heads the unprocessed suffix. Rewire one node while preserving access to the suffix.
210 карток
Coding Interview Patterns Flashcards: Signals, Invariants & Complexity
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Картка 106
Питання
How can you remove the nth node from the end of a list in one pass?
Відповідь
Start lead at the head and lag at a dummy head. Advance lead n nodes, then move both until lead is null; remove lag.next. Reject invalid n according to the input contract.
Картка 107
Питання
Which tree traversal naturally computes a value that depends on both children's results?
Відповідь
Postorder. Process left and right subtrees before combining their results at the parent.
Картка 108
Питання
How can you locate a cycle's entry after Floyd's two-speed pointers meet?
Відповідь
Reset one pointer to the head and move both one step at a time. Their next meeting is the entry; this follows from the distances modulo the cycle length.
Картка 109
Питання
What is the difference between tree depth and tree height?
Відповідь
Depth measures distance from the root to a node; height measures the longest downward distance to a leaf. State whether distances count edges or nodes.
Картка 110
Питання
How can you merge two sorted linked lists using O(1) auxiliary node storage?
Відповідь
Relink the smaller current node onto the result tail, advancing that list. Attach the remaining suffix when one list ends; existing nodes are reused.
Картка 111
Питання
When is breadth-first traversal more natural than depth-first traversal on a tree?
Відповідь
When results are grouped by depth or you need the nearest qualifying node by edge count. A queue processes one distance layer before the next.
Картка 112
Питання
Why must linked-list intersection compare node identity rather than node value?
Відповідь
Intersection means sharing the same node object and suffix. Separate nodes can hold equal values without the lists intersecting.
Картка 113
Питання
Why is checking only immediate children insufficient to validate a binary search tree?
Відповідь
A descendant can satisfy its parent yet violate an ancestor's constraint. Carry inherited lower and upper bounds, with an explicit duplicate policy.
Картка 114
Питання
What is the lowest common ancestor of two nodes in a rooted tree?
Відповідь
The deepest node that is an ancestor of both, allowing a node to be its own ancestor.
Картка 115
Питання
What property makes inorder traversal useful in a binary search tree?
Відповідь
It visits keys in sorted order under the tree's duplicate policy. In a strict BST, each visited key must be greater than the previous one.
Картка 116
Питання
Why must maximum tree-path sum separate its returned value from its global candidate?
Відповідь
The parent can extend only one downward branch. A complete path considered at the current node may join both children, but that fork cannot be extended upward.
Картка 117
Питання
How can two pointers find the intersection of two acyclic singly linked lists without measuring lengths?
Відповідь
After reaching a list's end, switch that pointer to the other head. Each traverses both lengths, so they meet at the shared node or at null.
Картка 118
Питання
What extra information makes preorder serialization unambiguous for an arbitrary binary tree?
Відповідь
Explicit null-child markers or another equivalent shape encoding. Values in preorder alone do not determine the structure.
Картка 119
Питання
Why can repeated subtree-height calculations make a tree-balance check O(n²)?
Відповідь
The same descendants may be scanned from many ancestors. Return height and balance together in one postorder traversal to visit each node once.
Картка 120
Питання
How does BST ordering guide a lowest-common-ancestor search for two existing distinct keys?
Відповідь
Move left if both keys are smaller and right if both are larger. The first split, or a node matching one key, is their lowest common ancestor.
Картка 121
Питання
What runtime does a search in an ordinary unbalanced BST guarantee?
Відповідь
O(h), where h is its height, and O(n) in the worst case. Logarithmic search requires a balance guarantee or a stated expected-shape assumption.
Картка 122
Питання
When computing a root-to-leaf path sum, why is reaching a null child insufficient for success?
Відповідь
A valid endpoint must be a leaf with no children. A missing child beside an existing child does not finish a root-to-leaf path.
Картка 123
Питання
What does a min-heap guarantee about its root and children?
Відповідь
The root is a minimum element, and each parent is no larger than its children. The whole array representation is not sorted.
Картка 124
Питання
A stream needs the k largest values seen so far. Which heap should you maintain?
Відповідь
A min-heap of at most k values. Its root is the smallest retained value, so a larger arrival can replace it.
Картка 125
Питання
What shared structure does a trie store?
Відповідь
Prefixes of keys. Following one edge per symbol reaches a key's prefix node, while a terminal marker distinguishes a complete stored key.
Картка 126
Питання
What are the usual binary-heap costs for peek, insertion, and root removal?
Відповідь
Peek is O(1); insertion and root removal are O(log n). Moving a changed element along one root-to-leaf path restores heap order.
Картка 127
Питання
How can a heap merge k sorted input streams?
Відповідь
Keep each nonempty stream's next element in a min-heap. Emit the minimum and replace it with that stream's next item. With k streams and N total elements, O(k + N log k) time includes initialization for k ≥ 2.
Картка 128
Питання
Why is bottom-up heap construction O(n), not O(n log n)?
Відповідь
Most nodes are near the leaves and can move only a short distance. Summing each node's possible sift-down work across all heights is linear.
Картка 129
Питання
How do two heaps support a running median?
Відповідь
Keep the lower half in a max-heap and the upper half in a min-heap, with sizes differing by at most one and every lower value no greater than every upper value.
Картка 130
Питання
What is a trie's lookup cost for a key of length L?
Відповідь
O(L) when each child transition is O(1). Child maps or ordered child containers can change that transition cost; space depends on stored prefixes and representation.
Картка 131
Питання
Why does a trie node need a terminal marker even if it has children?
Відповідь
A stored key can be a prefix of another key. The marker distinguishes a complete word from a prefix that merely leads to longer words.
Картка 132
Питання
When does sorting make more sense than a top-k heap?
Відповідь
When you need the entire sorted order or k is close to n and a full sort is acceptable. A heap's advantage is strongest when only a small retained subset is needed.
Картка 133
Питання
Why does a priority queue not by itself support efficient arbitrary deletion?
Відповідь
The heap efficiently exposes only its root. Removing another item needs its position, an indexed-heap design, or a lazy-deletion scheme with cleanup.
Картка 134
Питання
What output cost remains after a trie reaches a requested prefix?
Відповідь
Enumerating matching completions still costs time proportional to the visited subtree and emitted text. Prefix lookup does not make all autocomplete results free.
Картка 135
Питання
How can a priority queue break tied priorities without comparing the payloads?
Відповідь
Attach a unique increasing sequence number and compare priority first, then sequence number. Equal-priority items can then leave in insertion order without requiring an order on their payloads.
Картка 136
Питання
Why can a trie use more memory than a hash set of complete strings?
Відповідь
Nodes and child containers have overhead, especially for sparse branches. Shared prefixes save repeated symbols but do not guarantee a smaller representation.
Картка 137
Питання
What should graph modeling identify before choosing a traversal?
Відповідь
The states as vertices and legal transitions as edges, including direction and cost. A grid cell, word, or puzzle configuration can be a vertex.
Картка 138
Питання
When does ordinary BFS find a shortest path?
Відповідь
When every edge has the same nonnegative cost, including the unweighted case. Processing vertices by distance layer makes first discovery a shortest-edge-count path.
Картка 139
Питання
What is the space cost of an adjacency list compared with an adjacency matrix?
Відповідь
A list uses O(V + E) space; a matrix uses O(V²). A matrix gives constant-time edge lookup, while lists efficiently enumerate actual neighbors.
Картка 140
Питання
Which pattern finds all vertices reachable from a start vertex?
Відповідь
DFS or BFS with a visited set. Each reachable vertex and edge is processed a bounded number of times with adjacency lists.
Картка 141
Питання
What role does a parent map play in shortest-path traversal?
Відповідь
It records the predecessor used to reach each state. After reaching the target, follow parents backward and reverse the sequence to recover a path.
Картка 142
Питання
Why should BFS mark a vertex visited when enqueuing it?
Відповідь
To prevent several parents from adding it before its first removal. First enqueue already fixes its distance in an unweighted graph.
Картка 143
Питання
When is Dijkstra's algorithm appropriate?
Відповідь
For shortest paths with nonnegative edge weights. Its greedy finalization relies on no later path reducing a settled distance through a negative edge.
Картка 144
Питання
How can a grid traversal avoid confusing physical cells with full search states?
Відповідь
Include all information that changes future moves in the visited key, such as remaining obstacle removals or collected keys. Position alone may merge different states.
Картка 145
Питання
How do you detect a directed cycle with DFS?
Відповідь
Track unvisited, active, and finished vertices. An edge to an active vertex closes a cycle on the current recursion path.
Картка 146
Питання
Why can DFS with a visited set fail to find a shortest unweighted path?
Відповідь
Its first discovered route may follow a deep detour. DFS reachability order is not distance order; BFS provides that guarantee.
Картка 147
Питання
What does a topological ordering guarantee?
Відповідь
For every directed edge u → v, u appears before v. Such an ordering exists exactly when the directed graph is acyclic.
Картка 148
Питання
Why should stale priority-queue entries be skipped in a common Dijkstra implementation?
Відповідь
A vertex can receive a better distance after an older entry was pushed. Skip an entry whose stored distance differs from the current best distance.
Картка 149
Питання
For an undirected simple graph, why does DFS ignore the edge back to its parent when detecting cycles?
Відповідь
That edge is the same tree edge traversed in reverse, not a new cycle. A different already-visited neighbor indicates a cycle.
Картка 150
Питання
What does union-find answer efficiently?
Відповідь
Whether elements belong to the same connected component while components are merged. It does not store the actual connecting paths.
Картка 151
Питання
How does Kahn's algorithm build a topological ordering?
Відповідь
Enqueue all zero-indegree vertices, repeatedly remove one, and decrement its outgoing neighbors' indegrees. Enqueue each neighbor when its indegree becomes zero.
Картка 152
Питання
Which shortest-path algorithm handles edges weighted only 0 or 1 without a heap?
Відповідь
0–1 BFS with a deque. Push a relaxed zero-cost neighbor to the front and a one-cost neighbor to the back, preserving distance order.
Картка 153
Питання
Why is a boolean visited flag usually wrong for Dijkstra at first enqueue?
Відповідь
The first tentative distance need not be the shortest. Allow improvements; a vertex becomes settled when its smallest current distance is removed from the queue.
Картка 154
Питання
How do path compression and union by size or rank affect union-find complexity?
Відповідь
Together they give O(α(n)) amortized time per operation, where α is the inverse Ackermann function. The bound is effectively tiny for practical input sizes.
Картка 155
Питання
What does processing fewer than V vertices in Kahn's algorithm reveal?
Відповідь
A directed cycle remains. No vertex in the cyclic remainder can reach indegree zero after all removable dependencies are processed.
Картка 156
Питання
How can BFS compute distance from every grid cell to the nearest source?
Відповідь
Initialize the queue with all sources at distance zero. This multi-source BFS expands the nearest-source distance layers together.
Картка 157
Питання
Why can a topological ordering be nonunique?
Відповідь
Several vertices may currently have no remaining prerequisites. Choosing them in different orders can produce different valid orderings.
Картка 158
Питання
What happens when union-find receives an edge whose endpoints already share a representative?
Відповідь
The edge connects vertices already in one component. In incremental construction of an undirected forest, adding it creates a cycle.
Картка 159
Питання
Which algorithm can handle negative edge weights and detect a reachable negative cycle?
Відповідь
Bellman–Ford. Repeatedly relax all edges; an improvement after V − 1 full rounds indicates a negative cycle reachable from the source.
Картка 160
Питання
Why does traversal need an outer loop to count every connected component of an undirected graph?
Відповідь
One traversal reaches only one component. Start another traversal from each still-unvisited vertex and increment the component count.
Картка 161
Питання
How can topological order simplify shortest paths in a weighted DAG?
Відповідь
Relax each vertex's outgoing edges in topological order. Every predecessor is processed first, so negative weights are allowed and total time is O(V + E).
Картка 162
Питання
When should you use BFS or DFS instead of union-find for connectivity?
Відповідь
When the graph is static and you need traversal details such as paths or component members. Union-find is especially useful for repeated incremental edge additions and connectivity queries.
Картка 163
Питання
What is the difference between a minimum spanning tree and a shortest-path tree?
Відповідь
A minimum spanning tree minimizes total connecting edge weight. A shortest-path tree preserves shortest routes from a chosen source; neither objective implies the other.
Картка 164
Питання
Why can stopping at the first meeting be unsafe in bidirectional BFS with arbitrary node-by-node expansion?
Відповідь
A first meeting under an arbitrary expansion order may not minimize the combined distances. Use a layer-based stopping rule that accounts for both search depths.
Картка 165
Питання
Which signal suggests backtracking rather than a single greedy choice?
Відповідь
The task asks for all valid arrangements, or choices must be tried and undone because no safe local choice is known. Build a partial candidate and explore legal extensions.
Картка 166
Питання
What belongs in a backtracking state?
Відповідь
Enough information to determine legal next choices and recognize completion, such as the current position, chosen items, and remaining constraints.
Картка 167
Питання
What makes a pruning condition safe?
Відповідь
It proves that no completion of the current partial state can satisfy the goal or improve the required objective. A guess about likely failure is insufficient.
Картка 168
Питання
How do combinations differ from permutations during generation?
Відповідь
Combinations ignore order, so restrict future choices to later positions. Permutations care about order, so track which positions are already used.
Картка 169
Питання
What should be true after a backtracking recursive call returns?
Відповідь
The caller's mutable search state is restored exactly to its pre-choice state. Undo additions and constraint updates before trying a sibling choice.
Картка 170
Питання
When generating unique subsets from sorted values, how do you skip duplicates safely?
Відповідь
At one recursion depth, skip a value equal to the previous sibling candidate. Still allow equal values at deeper levels when the input provides multiple copies.
Картка 171
Питання
Why can backtracking output alone require exponential time?
Відповідь
A set with n distinct elements has 2^n subsets. Explicitly listing all subsets cannot be polynomial in n; copying their contents adds further cost.
Картка 172
Питання
Why should a completed mutable candidate usually be copied before saving it?
Відповідь
Later backtracking steps will modify the working candidate. Saving only a reference can make all recorded answers reflect subsequent changes.
Картка 173
Питання
What is the main risk of memoizing backtracking solely by the current index?
Відповідь
Different histories can leave different remaining choices or constraints. The memo key must include every part of the state that affects future results.
Картка 174
Питання
For selecting the most nonoverlapping intervals, which greedy choice is justified?
Відповідь
Choose the available interval with the earliest finishing time, then continue with compatible intervals. This leaves at least as much room for the remaining selections.
Картка 175
Питання
What is the difference between greedy choice and dynamic programming?
Відповідь
Greedy commits to a choice proven safe without exploring all alternatives. DP evaluates and combines subproblem alternatives when that local commitment is not justified.
Картка 176
Питання
Why does choosing the largest coin repeatedly fail for some coin systems?
Відповідь
The locally largest coin can leave an expensive remainder. With denominations 1, 3, 4 and amount 6, greedy uses 4 + 1 + 1, while 3 + 3 uses fewer coins.
Картка 177
Питання
How does a farthest-reachable frontier solve reachability in a nonnegative jump-length array?
Відповідь
Scan positions no farther than the current frontier and extend it with each reachable index plus its jump length. If the next position lies beyond the frontier, progress is impossible.
Картка 178
Питання
Why does choosing the shortest interval not always maximize the number of nonoverlapping intervals?
Відповідь
A short interval can cross the boundary between two compatible intervals and block both. Duration alone does not measure the future scheduling space it consumes.
Картка 179
Питання
What must be proved before pruning a combination-sum branch because its sum exceeds the target?
Відповідь
Remaining choices cannot reduce the sum. The pruning is safe for nonnegative additions under the stated goal, but negative numbers can make it invalid.
Картка 180
Питання
How does branch and bound differ from ordinary feasibility pruning?
Відповідь
It uses a bound on the best objective reachable from a partial state. Prune only when that bound cannot beat the best complete answer already found.
Картка 181
Питання
What question distinguishes a greedy proof from evidence that a heuristic often works?
Відповідь
Can every discarded alternative be ruled out for all valid inputs? Examples and benchmarks support a heuristic, but do not establish the safe-choice property.
Картка 182
Питання
Why is earliest-finish interval scheduling insufficient when intervals have different rewards?
Відповідь
Maximizing count and maximizing reward are different objectives. A single high-reward interval can beat several low-reward intervals, so weighted scheduling needs more information.
Картка 183
Питання
Which combination of properties makes dynamic programming promising?
Відповідь
Repeated subproblems and a recurrence that combines their results. Define a state whose answer is independent of the path used to reach it.
Картка 184
Питання
What should a DP state definition say before you write a recurrence?
Відповідь
Exactly what one table entry means, including its input boundary and any remaining resource or constraint. Ambiguous states lead to mismatched transitions.
Картка 185
Питання
How do top-down memoization and bottom-up tabulation differ?
Відповідь
Memoization computes states on demand through calls and caches them. Tabulation processes states in an explicit dependency order, often avoiding recursion overhead.
Картка 186
Питання
What determines the runtime of a DP with a finite state table?
Відповідь
The number of states actually evaluated times the work per state, plus preprocessing and output reconstruction. Count transitions rather than just table dimensions.
Картка 187
Питання
Why are base cases part of a DP's meaning rather than convenient initial values?
Відповідь
They encode valid empty or smallest subproblems. A wrong base value can invent impossible solutions or remove legitimate ones from every later transition.
Картка 188
Питання
For 0/1 knapsack compressed to one capacity array, why iterate capacities downward?
Відповідь
Each item must be used at most once. Descending order reads the previous item's state instead of reusing an update made for the current item.
Картка 189
Питання
When can a DP table be compressed to a few rows or variables?
Відповідь
When future states depend only on a bounded slice of earlier states. Keep those dependencies until their last use; reconstruction may need additional storage.
Картка 190
Питання
What recurrence models choosing nonadjacent values for maximum sum?
Відповідь
At each position, compare skipping it with taking it plus the best result before its neighbor. The base cases must specify whether choosing nothing is allowed.
Картка 191
Питання
What DP state counts paths through a blocked grid when moves are only right or down?
Відповідь
The number of ways to reach each cell from above or from the left. Blocked cells contribute zero; initialize an unblocked starting cell to one.
Картка 192
Питання
For unbounded knapsack, why can capacities run upward within an item's pass?
Відповідь
Reusing the current item's updated smaller-capacity result is allowed. Ascending order lets that item contribute more than once.
Картка 193
Питання
How can loop order change coin-change counting from combinations to ordered sequences?
Відповідь
Processing coin types outside amounts builds combinations without ordering them. Processing amounts outside all coin choices counts different last-coin sequences separately.
Картка 194
Питання
What is the key distinction between longest common subsequence and longest common substring?
Відповідь
A subsequence may skip characters; a substring must stay contiguous. Their DP transitions differ because a substring match cannot carry through a mismatch.
Картка 195
Питання
Why is O(nW) knapsack called pseudopolynomial?
Відповідь
It is polynomial in the numeric capacity W, but W can be exponential in the number of bits used to encode it. It is not polynomial in input bit length.
Картка 196
Питання
For longest increasing subsequence, what does tails[length − 1] represent in the O(n log n) method?
Відповідь
The smallest possible final value of an increasing subsequence of that length among processed values. The tails array itself need not be one actual subsequence.
Картка 197
Питання
Why do counting and minimization DPs use different unreachable-state values?
Відповідь
A count uses zero ways. A minimization state needs an explicit unreachable marker or infinity so an impossible predecessor cannot look like a cheap solution.
Картка 198
Питання
What state supports edit distance between two strings?
Відповідь
The minimum edits needed to transform one prefix into the other. Transitions account for insertion, deletion, and replacement or a matching final character.
Картка 199
Питання
How can a DP recover one chosen solution instead of only its score?
Відповідь
Store predecessor or choice information, or recompute choices from the full table. Walk backward from the final state to reconstruct the selected decisions.
Картка 200
Питання
Why must an LIS implementation choose its binary-search boundary according to strictness?
Відповідь
For a strictly increasing subsequence, replace the first tail at least equal to the value. A nondecreasing subsequence instead uses the first strictly greater tail.
Картка 201
Питання
What operation tests whether bit i of a nonnegative integer mask is set?
Відповідь
Check whether mask AND (1 shifted left by i) is nonzero. Ensure i is inside the integer representation's supported bit range.
Картка 202
Питання
Why does XOR recover a unique value when every other value occurs exactly twice?
Відповідь
Equal values cancel because x XOR x = 0, and XOR is associative and commutative. XORing all values leaves the single unpaired value.
Картка 203
Питання
What does x AND (x − 1) do for a positive integer x?
Відповідь
It clears x's lowest set bit. Repeating it counts set bits in time proportional to the number of set bits.
Картка 204
Питання
When does a bitmask make a useful DP state?
Відповідь
When a small set of items is either included or excluded and future choices depend on that subset. n items give 2^n possible masks, so n must be small.
Картка 205
Питання
How do you set a bit and clear a bit without changing the others?
Відповідь
Set bit i with mask OR (1 shifted left by i). Clear it with mask AND the bitwise complement of that single-bit mask, respecting the chosen word width.
Картка 206
Питання
What makes a memoized recurrence invalid when it depends on mutable global state omitted from the key?
Відповідь
The same key can have different answers under different global conditions. Include the relevant state in the key or remove that dependency.
Картка 207
Питання
What condition recognizes a power of two among integers?
Відповідь
x > 0 and x AND (x − 1) = 0. The positivity check excludes zero, which also makes the bitwise expression zero.
Картка 208
Питання
Why does bitwise complement need a width convention in language-agnostic reasoning?
Відповідь
Complement flips all bits in the representation. Fixed-width and arbitrary-precision signed integers can produce different-looking values; mask to the intended width when necessary.
Картка 209
Питання
How can two unique values be recovered when every other value occurs twice?
Відповідь
XOR all values, choose a set bit in that nonzero result, and partition by that bit. XOR within each group; the two unique values fall into different groups.
Картка 210
Питання
Why is memoization alone insufficient to handle cyclic state dependencies?
Відповідь
A call may revisit an unfinished state before any value is cached. Use cycle handling or a problem-specific iterative method; ordinary DAG-style DP assumes an acyclic dependency order.
210 карток
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