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