C++     Medium     Stack     String    

Problem Statement:

You are given a string s and an integer k, a k duplicate removal consists of choosing k adjacent and equal letters from s and removing them, causing the left and the right side of the deleted substring to concatenate together.

We repeatedly make k duplicate removals on s until we no longer can.

Return the final string after all such duplicate removals have been made. It is guaranteed that the answer is unique.

 

Example 1:

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

Example 2:

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"

Example 3:

Input: s = "pbbcggttciiippooaais", k = 2
Output: "ps"

 

Constraints:

  • 1 <= s.length <= 105
  • 2 <= k <= 104
  • s only contains lowercase English letters.

Solution:

Maintain a stack of (char, frequency) pair. Here is how it works for a test case.

Input:

 
s = "deeedbbcccbdaa", k = 3

Dry run:

First column is where you are at. Second column is the stack at that time

 
d : [['d', 1]] 
e : [['d', 1], ['e', 1]] 
e : [['d', 1], ['e', 2]] 
e : [['d', 1]] #You popped here
d : [['d', 2]] 
b : [['d', 2], ['b', 1]] 
b : [['d', 2], ['b', 2]] 
c : [['d', 2], ['b', 2], ['c', 1]] 
c : [['d', 2], ['b', 2], ['c', 2]] 
c : [['d', 2], ['b', 2]] #You popped here
b : [['d', 2]] #You popped here
d : [] #You popped here
a : [['a', 1]] 
a : [['a', 2]]

Answer:

Just join the elements in stack:

 
"aa"

Now we are ready to code!

Code:

 
class Solution {
public:
    string removeDuplicates(string s, int k) {
        stack<pair<char,int>> H;
        for (char ch: s)
        {
            if (H.empty() || H.top().first!=ch) H.push({ch,1});
            else
            {
                H.top().second++;
                if (H.top().second==k) H.pop();
            }
        }
        string t="";
        while (!H.empty())
        {
            auto p=H.top(); 
            t+=string(p.second,p.first);
            H.pop();
        }
        reverse(t.begin(),t.end());
        return t;
    }
};