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