Approach: Stack

def removeDuplicates(s):
    st = []
    for c in s:
        if st and st[-1] == c: st.pop()
        else: st.append(c)
    return "".join(st)

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

Full Solution — all languages

class Solution:
    def removeDuplicates(self, s):
        st = []
        for c in s:
            if st and st[-1] == c:
                st.pop()
            else:
                st.append(c)
        return "".join(st)

import java.util.*;
class Solution {
    public String removeDuplicates(String s) {
        StringBuilder sb = new StringBuilder();
        for (char c : s.toCharArray()) {
            if (sb.length() > 0 && sb.charAt(sb.length() - 1) == c) sb.deleteCharAt(sb.length() - 1);
            else sb.append(c);
        }
        return sb.toString();
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: string removeDuplicates(string s) {
        string st;
        for (char c : s) {
            if (!st.empty() && st.back() == c) st.pop_back();
            else st.push_back(c);
        }
        return st;
    }
};

/**
 * @param {string} s
 * @return {string}
 */ var removeDuplicates = function(s) {
    const st = [];
    for (const c of s) {
        if (st.length && st[st.length - 1] === c) st.pop();
        else st.push(c);
    }
    return st.join('');
};

Remove All Adjacent Duplicates In String

Example 1

Example 2

Constraints