Remove Duplicates from Sorted Array

Remove duplicates from sorted array removes duplicates in place, returning the count of unique elements. The array is sorted, so duplicates are always adjacent — which is what makes a single pass possible. The problem's contract is worth reading carefully. It asks for the count, and only requires the first k positions to hold the unique values. Whatever sits beyond position k is ignored, so nothing needs shifting, erasing, or resizing. That frees the solution to use two pointers with different roles: - A write pointer marks where the next unique value belongs, while a read pointer scans forward looking for values not yet written. The write pointer starts at 1, since the first element is always unique and already in place. For each element, compare against the value at write − 1 — the last unique value written. If they differ, write it and advance. Comparing against nums[write − 1] rather than nums[read − 1] is the detail that matters. The last written value is the correct reference, and comparing with the previous read position breaks once the two pointers diverge. The write pointer's final value is the count, which is exactly what the function returns. An empty array returns 0, and a single element returns 1 — both fall out if the write pointer starts at 1 and the loop begins at index 1, since the loop simply does not run. Remove Duplicates from Sorted Array II allows each value twice, and adapts by comparing against nums[write − 2] instead. One pass with two indices gives O(n) time and O(1) space.

The read/write two-pointer: one index scans every element, a slower one marks where the next kept value goes. Because the array is sorted, duplicates are adjacent and a single comparison against the last kept value decides. Any in-place "remove or compact" problem uses this same slow-write, fast-read split.