Remove All Adjacent Duplicates in String II

Last Updated : 10 Jun, 2024

Given a string s and an integer k, the task is to repeatedly delete k adjacent duplicates till no deletions are possible and then return the final string. On deletion of k adjacent duplicates, the left and right sides of the deleted substring is concatenated together.

Examples:

Input: s = "abcd", k = 2
Output: "abcd"
Explanation: There's nothing to delete.

Input: s = "deeedbbcccbdaa", k = 3
Output: "aa"
Explanation:

First delete "eee" and "ccc", get "ddbbbdaa".
Then delete "bbb", get "dddaa".
Finally delete "ddd", get "aa".

Approach: To solve the problem, follow the below idea:

To solve this problem, we can use a stack to keep track of characters and their counts. The idea is to use the stack to store pairs of characters and their consecutive counts. Whenever we encounter k consecutive characters, we remove them from the stack.

Step-by-step algorithm:

Initialize a stack that will store pairs of characters and their consecutive counts.
Iterate through the string:
- For each character, check if the stack is not empty and the top character of the stack is the same as the current character.
- If they are the same, increment the count.
- If the count reaches k, pop the stack.
- If they are not the same, push the current character with a count of 1 onto the stack.
Construct the resulting string from the stack.
Return the resulting string.

Below is the implementation of the algorithm:

C++

#include <bits/stdc++.h>
using namespace std;

string removeDuplicates(string s, int k)
{
      // stack to store the character with its count
    stack<pair<char, int> > st;

    for (char c : s) {
          // If the character is same as character at top, increase the count of top character
        if (!st.empty() && st.top().first == c) {
            st.top().second++;
            if (st.top().second == k) {
                st.pop();
            }
        }
          // If the character is different from the character at top, push the character to the top
        else {
            st.push({ c, 1 });
        }
    }

    string result = "";
  
     // Construct the result string
    while (!st.empty()) {
        result.append(st.top().second, st.top().first);
        st.pop();
    }
    reverse(result.begin(), result.end());
    return result;
}

int main()
{
    // Example 1
    string s1 = "abcd";
    int k1 = 2;
    cout << removeDuplicates(s1, k1)
         << endl; // Output: "abcd"

    return 0;
}

Java

import java.util.*;

public class RemoveDuplicates {

    public static String removeDuplicates(String s, int k) {
        // Stack to store the character with its count
        Deque<Pair> stack = new ArrayDeque<>();

        for (char c : s.toCharArray()) {
            // If the character is same as the character at top, increase the count of top character
            if (!stack.isEmpty() && stack.peek().character == c) {
                stack.peek().count++;
                if (stack.peek().count == k) {
                    stack.pop();
                }
            }
            // If the character is different from the character at top, push the character to the top
            else {
                stack.push(new Pair(c, 1));
            }
        }

        StringBuilder result = new StringBuilder();

        // Construct the result string
        while (!stack.isEmpty()) {
            Pair p = stack.pop();
            for (int i = 0; i < p.count; i++) {
                result.append(p.character);
            }
        }

        return result.reverse().toString();
    }

    public static void main(String[] args) {
        // Example 1
        String s1 = "abcd";
        int k1 = 2;
        System.out.println(removeDuplicates(s1, k1)); // Output: "abcd"
    }
}

class Pair {
    char character;
    int count;

    Pair(char character, int count) {
        this.character = character;
        this.count = count;
    }
}

// This code is contributed by Shivam Gupta

Output

abcd

Time Complexity: O(n), where n is the length of the string.
Auxiliary Space: O(n)

Related Articles:

Remove All Duplicates from a Given String in Python

codernitnucy

Improve

Article Tags :

Practice Tags :

Remove All Adjacent Duplicates in String II

Similar Reads

Thank You!

What kind of Experience do you want to share?