arrays

The most common screen opener, and interviewers use it to check whether hashmap thinking is reflexive. Here's the one-pass answer, the narration that earns points, and the duplicate-handling edge case most candidates fumble.

The follow-up interviewers reach for when merge-intervals goes too smoothly. The three-phase scan is elegant, but only if you've internalized the overlap condition most candidates have to re-derive under pressure.

Looks like a sliding-window problem; isn't one. The prefix-sum + hashmap trick that solves it is the same idea behind sessionization and cumulative-metrics questions later in the loop: learn it once, reuse it three times.

Everyone knows the sort + two-pointer outline. The interview is actually about duplicate handling: three separate skip conditions that candidates routinely botch live. Here's the clean version and the narration that goes with it.

C3.ai runs the most LeetCode-classic FDE loop, and candidates report the same published set for years: Trapping Rain Water on the phone screen, Daily Temperatures in the tech screen. Both optimal solutions, plus the one-line insight that unlocks each.

The classic linear DP. At each house you either skip it and keep the best so far, or take it and add the best from two back. The signal is collapsing the table to two rolling variables and then handling the circular follow-up cleanly.

Build the output from prefix products on a left-to-right pass, then fold in suffix products on a right-to-left pass using the output array itself as scratch. O(n) time, O(1) extra space. The interesting part is why division is banned and what it costs when the input has zeros.

The O(log(min(m,n))) answer binary-searches a partition on the shorter array so the left halves of both arrays together hold exactly half the elements. The whole problem is one invariant plus careful boundary handling for empty sides and odd versus even totals.

A sorted array got rotated at an unknown pivot and you still have to find a target in O(log n). The trick is deciding which half is sorted at every step. Here's the clean invariant, and the duplicates follow-up that quietly breaks the log-n promise.