Formatted question description: https://leetcode.ca/all/1209.html

1209. Remove All Adjacent Duplicates in String II (Medium)

Given a string s, 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 <= 10^5
  • 2 <= k <= 10^4
  • s only contains lower case English letters.

Related Topics:
Stack

Similar Questions:

Solution 1. Stack

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

// Time: O(N)
// Space: O(N)
class Solution {
public:
    string removeDuplicates(string s, int k) {
        vector<pair<char, int>> st;
        for (char c : s) {
            if (st.empty() || st.back().first != c) st.emplace_back(c, 1);
            else if (++st.back().second == k) st.pop_back();
        }
        string ans;
        for (int i = 0; i < st.size(); ++i) {
            for (int j = 0; j < st[i].second; ++j) ans += st[i].first;
        }
        return ans;
    }
};

Java

class Solution {
    public String removeDuplicates(String s, int k) {
        StringBuffer sb = new StringBuffer();
        List<Integer> duplicatesCounts = new ArrayList<Integer>();
        int length = s.length();
        for (int i = 0; i < length; i++) {
            int curLength = sb.length();
            char c = s.charAt(i);
            int curCount = 1;
            if (curLength > 0 && c == sb.charAt(curLength - 1))
                curCount = duplicatesCounts.get(curLength - 1) + 1;
            if (curCount == k) {
                int startIndex = curLength - k + 1;
                sb.delete(startIndex, curLength);
                for (int j = curLength - 1; j >= startIndex; j--)
                    duplicatesCounts.remove(j);
            } else {
                sb.append(c);
                duplicatesCounts.add(curCount);
            }
        }
        return sb.toString();
    }
}

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