Instead of checking every possible pair, we use the properties of the problem to make a greedy decision at each step, resulting in an elegant O(n) solution.

The key insight is realizing that after sorting the array, we don’t need three nested loops anymore. Instead: Fix one number., use two pointers to search for the other two, skip duplicates carefully and stop early once the fixed number becomes positive.

A sorted array often hints at using the two-pointer technique. Instead of checking every possible pair, eliminate impossible candidates by moving pointers intelligently.

This is a classic interview problem because it combines string manipulation with the Two Pointer pattern while keeping both the time complexity O(n) and the space complexity O(1).

Given an integer array nums and an integer val, the task is to remove all occurrences of val in nums in-place. Additionally, we need to return the number of elements in nums that are not equal to val.

The problem at hand is to remove duplicates from a sorted integer array nums in-place, such that each unique element appears only once. Additionally, we need to return the number of unique elements present in the modified array.

You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively. Merge nums1 and nums2 into a single array sorted in non-decreasing order.