Approach: Two Pointers

def removeDuplicates(nums):
    k = 0
    for x in nums:
        if k == 0 or x != nums[k-1]:
            nums[k] = x; k += 1
    return k

Complexity: O(n) time, O(1) space.

Full Solution — all languages

class Solution:
    def removeDuplicates(self, nums):
        k = 0
        for x in nums:
            if k == 0 or x != nums[k - 1]:
                nums[k] = x
                k += 1
        return k

import java.util.*;
class Solution {
    public int removeDuplicates(int[] nums) {
        int k = 0;
        for (int x : nums) {
            if (k == 0 || x != nums[k - 1]) {
                nums[k] = x;
                k++;
            }
        }
        return k;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int removeDuplicates(vector<int>& nums) {
        int k = 0;
        for (int x : nums) {
            if (k == 0 || x != nums[k - 1]) {
                nums[k] = x;
                k++;
            }
        }
        return k;
    }
};

/**
 * @param {number[]} nums
 * @return {number}
 */ var removeDuplicates = function(nums) {
    let k = 0;
    for (const x of nums) {
        if (k === 0 || x !== nums[k - 1]) {
            nums[k] = x;
            k++;
        }
    }
    return k;
};

Remove Duplicates from Sorted Array

Example 1

Example 2

Constraints