Problem statement

You are given a string s and an integer k, a k duplicate removal consists of choosing k adjacent and the same 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 is 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 <= 10^5.

• 2 <= k <= 10^4.

• s only contains lower case English letters.

Solution: Strings of adjacent equal letters

Construct a stack of strings that has adjacent equal letters and perform the removal during building those strings.

Example 2

For s = "deeedbbcccbdaa" and k = 3:

• The first built string is "d".

• Then "eee" with exact length k, remove this string.

• The next character is 'd', which equals the last character of the last string "d", merge them together. The first string becomes "dd".

• The next string is "bb".

• Then "ccc" is removed.

• The next character 'b' is merged with the last string ("bb") to become "bbb" and be removed.

• The next character 'd' is merged with the last string ("dd") to become "ddd" and be removed.

• The remaining string is "aa".

Code

#include <iostream>
#include <vector>
using namespace std;
string removeDuplicates(string s, int k) {
    vector<string> stk;
    int i = 0;
    while (i < s.length()) {
        string a;   // to store adjacent equal letters        
        // perform the merge
        if (!stk.empty() && s[i] == stk.back().back()) {
            a = move(stk.back());
            stk.pop_back();
        }
        int j = i;
        while (j < s.length() && s[j] == s[i]) {
            a += s[j];
            // remove the k-duplicate
            if (a.length() == k) {
                a = "";
            }
            j++;
        }
        if (!a.empty()) {
            stk.push_back(a);
        }
        i = j;
    }
    s = "";
    for (auto& str : stk) {
        s += str;
    }
    return s;
}
int main() {
    cout << removeDuplicates("abcd", 2) << endl;
    cout << removeDuplicates("deeedbbcccbdaa", 3) << endl;
    cout << removeDuplicates("pbbcggttciiippooaais", 2) << endl;
}

abcd
aa
ps

Complexity

• Runtime: O(N), where N = s.length.

• Extra space: O(N).

Implementation notes

The data structure stk you might need to solve this problem is a stack. But here are the reasons you had better use std::vector:

• std::vector has also methods push_back(value) and pop_back() like the ones in stack.

• On the other hand, it is faster for a vector to perform the string concatenation at the end.

References

• https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/

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