two pointers

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 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.