InterviewGrind › Coding patterns
Two pointers, sliding windows, BFS, binary search on the answer and more: pick the right pattern for each problem, the skill that unlocks most coding interviews.
15 free questions from the app. Try each one before you open the answer.
Answer: A. Stack
Clue: nesting, where the most recent opener must close first (last in, first out). Push openers; on a closer, the top of the stack must be its match. The stack must be empty at the end.
Answer: A. Prefix sums: each query is prefix[j + 1] - prefix[i]
One O(n) pass makes every query O(1). The 2D version (summed-area table) answers rectangle sums in a grid.
Answer: A. Sort by start, then merge each into the last kept interval
Clue: intervals plus "overlap". After sorting by start, an interval overlaps the previous merged one only if its start ≤ that one's end. O(n log n) for the sort, then one pass.
Answer: A. Flood fill (DFS or BFS) from each unvisited land cell
Scan the grid; each time you hit unvisited land, count an island and flood-fill everything connected to it so you don't count it again. O(rows × cols). Union-find also works.
Answer: B. Two pointers from both ends
Clue: sorted input plus a pair condition. If the sum is too small, move the left pointer right; if too big, move the right pointer left. O(n) time, O(1) space.
Answer: A. Fast and slow pointers
Move one pointer one step and the other two steps at a time. When the fast one reaches the end, the slow one is in the middle. O(n) time, O(1) space.
Answer: A. Myth
Myth. Sorted input is a strong hint, but it often points to two pointers (pair sums, removing duplicates, merging) rather than binary search. Ask what the sortedness lets you skip.
Answer: A. Backtracking: include or skip each element
Clue: "all combinations / subsets / permutations". Each element is a choose-or-skip decision, giving 2ⁿ subsets. Iterating a bitmask from 0 to 2ⁿ - 1 is an equivalent alternative.
Answer: A. Sliding window
Clue: "longest contiguous substring" with a constraint you can update incrementally. Grow the right edge, shrink the left edge while there are more than K distinct characters, and track the best width. O(n) time, O(K) space.
Answer: B. Min-heap of size K
Clue: "top K" over a stream. Keep only the K largest in a min-heap; its root is the Kth largest. Each insert is O(log K), and space is O(K).
Answer: B. Monotonic stack
Clue: "next greater/smaller". Keep a stack of indices still waiting for an answer, in decreasing order of value. Each new item pops every smaller item and becomes their answer. Every index is pushed and popped once: O(n).
Answer: A. Union-find (disjoint set union)
Clue: "connected groups" with merges arriving incrementally. Each union merges two sets in nearly O(1) (with path compression and union by rank), and you track the group count as you go. BFS/DFS also works on a fixed graph but must re-run when edges are added.
Answer: A. Binary search, checking which half is sorted at each step
Clue: "sorted" + O(log n) required. At every step, compare nums[lo] with nums[mid] to find which half is in order, then decide whether the target lies in that half.
Answer: A. Pick the meeting that ends earliest, repeatedly
Sort by end time and take each meeting that starts after the last chosen one ends. Finishing earliest leaves the most room for the rest; an exchange argument proves it. O(n log n).
Answer: A. Count with a hash map, then keep a min-heap of size K
Clue: "top K" by some score. Counting is O(n); a size-K heap over the distinct words is O(m log K). Bucket sort by frequency (index = count) gets it to O(n) if asked for better.
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