DSA interview questions and answers

Maintain a window of length p with character counts

Sliding window: the window is valid while window_size minus the count of the most frequent character is at most k.

This is a window where the number of zeros is at most k

Scan left to right and keep a stack of indices with decreasing temperatures

Store values on one stack and running minima on a second stack (or pairs of value and min).

Iterate asteroids and use a stack of survivors.

Use a stack of (previous_string, multiplier) pairs: on '[', push the current buffer and k, on ']', pop and repeat the inner buffer.

Keep a buffer, head index, count (or head and tail), and capacity.

Maintain a frequency map and a queue of candidate characters (or indices).

Multi-source BFS: enqueue every rotten orange at time 0, then spread to adjacent fresh oranges level by level.

Use a dummy head and append the smaller current node from either list, then attach any leftover tail.

Walk both lists with a carry, creating a new node for (sum % 10) and carrying sum // 10.

Find the middle with slow/fast, reverse the second half, compare pairwise, then optionally restore.

Two pointers walk both lists and switch to the other head on null

Split at the middle, reverse the second half, then weave nodes from the two halves.

The diameter is the max of left_height + right_height over every node (in edges).

Recurse: if the current node is p or q, return it

DFS subtracting the node value, succeeding when you hit a leaf and remaining target is zero.

A bottom-up DFS returns height or a sentinel if any subtree differs by more than one.

Level-order BFS and take the last node of each level, or DFS root-right-left and record the first node seen at each depth.