two pointers

Sliding window is one of the highest-yield coding patterns, and this is its canonical problem. The real signal is spotting when a window collapses an O(n squared) scan into one linear pass, and holding the invariant cleanly.

A classic that pays off the two-pointer insight over brute force. Interviewers look for realizing that water at each position is capped by the min of the max walls on each side, then computing it in a single pass. The answer follows.

Minimum Window Substring is the hard sliding-window problem that probes expand-and-contract with a character-count map. Interviewers look for the grow-to-valid then shrink-to-minimal pattern, verified in O(1) per step.

Sorting an array of three values in one pass probes the three-pointer partition (Dutch National Flag). Interviewers look for the low/mid/high invariant and knowing when not to advance mid. The answer follows.

This classic has a tidy O(n^2) expand-around-center solution that outdoes the naive O(n^3). Interviewers look for expanding from every center and handling odd and even lengths separately. The answer follows.

A row- and column-sorted matrix rewards a single insight: begin at a corner and drop a whole row or column each step. That converts an O(m*n) scan into O(m+n), and which corner you pick is not arbitrary.

Floyd's tortoise and hare finds a cycle in O(1) space, but what sets strong candidates apart is the second phase: a brief distance argument that identifies exactly where the loop starts. Here is the answer and the math.

Two lists that share a tail have to intersect at a single node, but aligning their lengths without counting is the elegant move interviewers look for. Here is the two-pointer trick that cancels the lengths out, plus the answer.

The two-pass length-then-delete solution works, but the interview asks for a single pass: a gap of N between two pointers, plus a dummy head that makes deleting the real head come out for free. Here is the clean answer.

The naive triple loop runs O(n cubed). Sort first, then fix one element and converge two pointers for O(n squared). The real test is deduplication: returning unique triplets without a set. Here is the clean template.