25. Remove Duplicates from Sorted Array

Because the array is sorted, duplicates are adjacent. Keep a slow write-pointer at the last unique value; advance a fast pointer, and whenever it finds a value different from the last kept one, write it forward.

Idea

Sorting groups equal values together, so a new unique value appears exactly when nums[fast] differs from the last kept value. A slow write-pointer k tracks how many uniques we've placed; we only write (and advance k) on a change.

Algorithm

1. If nums is empty, return []. Otherwise k = 1 (the first element is always kept).
2. For i from 1 to end: if nums[i] != nums[k-1], write nums[k] = nums[i] and k += 1.
3. The first k elements are the unique values; return nums[:k].

Worked example: [0,0,1,1,1,2,2,3,3,4]

• Keep 0. See 0 (dup). See 1 → write, k=2. 1,1 dups. See 2 → write, k=3. 2 dup. See 3 → write, k=4. 3 dup. See 4 → write, k=5.

Unique = [0,1,2,3,4] (k = 5).

Complexity

• Time: O(n).
• Space: O(1).

import java.util.Arrays;

class Solution {
    public int[] removeDuplicates(int[] nums) {
        if (nums.length == 0) {
            return new int[]{};
        }
        int k = 1;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] != nums[k - 1]) {
                nums[k] = nums[i];
                k++;
            }
        }
        return Arrays.copyOf(nums, k);
    }
}