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();
        }
    }
    
  • // 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 (auto &[c, cnt] : st) {
                while (cnt--) ans += c;
            }
            return ans;
        }
    };
    
  • # 1209. Remove All Adjacent Duplicates in String II
    # https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/
    
    class Solution:
        def removeDuplicates(self, s: str, k: int) -> str:
            stack = []
            
            for c in s:
                if stack and stack[-1][0] == c:
                    ch, cnt = stack.pop()
                    stack.append((ch, cnt + 1))
                else:
                    stack.append((c, 1))
                
                if stack and stack[-1][1] == k:
                    stack.pop()
            
            return "".join(ch * cnt for ch, cnt in stack)
    
    

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