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