When the array is sorted, the two-pointer sweep replaces the O(n^2) double loop with a single O(n) pass. The invariant is: every pair containing arr[lo] with an index < hi has already been considered, so if the current sum is too small, no smaller hi can help and we must move lo up. Symmetric reasoning lets us shrink hi when the sum is too large. Use this for twoSum on sorted input, finding pair-with-given-difference, or any problem that monotonically narrows on a sorted range.

Reversing in place uses the same two-pointer skeleton but swaps instead of moving conditionally. The loop runs n / 2 iterations, each doing one swap, so total work is O(n) with O(1) extra space. The empty-array case is handled by the loop condition without any special case. Anywhere you would reach for array.slice().reverse() you can use this version to save the allocation, which matters when the array is huge or you want to mutate in place anyway.

A slow-fast (or 'two pointers, same direction') sweep is the right tool for in-place compaction: filtering, deduplication of sorted arrays, and the classic 'move zeros to the end' problem. The read pointer scans every element, while write only advances when something is kept. Setting arr.length = write truncates the tail in O(1), avoiding a fresh allocation. This pattern also forms the inner loop of partition routines used by quicksort and Dutch-flag-style problems.