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[5/6] 1209. Remove All Adjacent Duplicates in String II

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작성일 2022.05.05 21:01

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[LeetCode 시즌 3] 2022년 5월 6일 문제입니다.

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

[Medium] 1209. Remove All Adjacent Duplicates in String II

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 lower case English letters.

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링크

https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/ 1602 회 연결

익명
작성일 2022.05.06 16:24

class Solution {
public:
    string removeDuplicates(string s, int k) {
        stack<pair<char, int>> st;
        string res;
        int n = s.size();
        
        for (int i = 0; i < n; ++i) {
            if (st.empty()) {
                st.push(pair<char, int>(s[i], 1));
            }
            else {
                auto top = st.top(); st.pop();
                if (top.first == s[i]) {
                    top.second = (top.second + 1) % k;
                    if (top.second > 0) {
                        st.push(top);
                    }
                }
                else {
                    st.push(top);
                    st.push(pair<char, int>(s[i], 1));
                }
            }
        }
        
        while (!st.empty()) {
            auto top = st.top(); st.pop();
            for (int i = 0; i < top.second; ++i) {
                res.push_back(top.first);
            }
        }
        reverse(res.begin(), res.end());
        return res;
    }
};

C++
Runtime: 26 ms, faster than 66.98% of C++ online submissions for Remove All Adjacent Duplicates in String II.
Memory Usage: 10.9 MB, less than 36.43% of C++ online submissions for Remove All Adjacent Duplicates in String II.
[code=C++]class Solution {
public:
    string removeDuplicates(string s, int k) {
        stack<pair<char, int>> st;
        string res;
        int n = s.size();
        
        for (int i = 0; i < n; ++i) {
            if (st.empty()) {
                st.push(pair<char, int>(s[i], 1));
            }
            else {
                auto top = st.top(); st.pop();
                if (top.first == s[i]) {
                    top.second = (top.second + 1) % k;
                    if (top.second > 0) {
                        st.push(top);
                    }
                }
                else {
                    st.push(top);
                    st.push(pair<char, int>(s[i], 1));
                }
            }
        }
        
        while (!st.empty()) {
            auto top = st.top(); st.pop();
            for (int i = 0; i < top.second; ++i) {
                res.push_back(top.first);
            }
        }
        reverse(res.begin(), res.end());
        return res;
    }
};[/code]

익명
작성일 2022.05.08 04:54

Runtime: 13 ms, faster than 96.55% of C++ online submissions for Remove All Adjacent Duplicates in String II.
Memory Usage: 10.3 MB, less than 63.87% of C++ online submissions for Remove All Adjacent Duplicates in String II.

class Solution {
public:
    string removeDuplicates(string s, int k) {
        char prev = 0U;
        int cnt = 0;
        string ret;
        stack<pair<char, int>> t;
        for(auto c: s) {
            ret.push_back(c);
            if (c != prev) {
                t.emplace(prev, cnt);
                prev = c, cnt = 1;
            } else if (++cnt == k) {
                ret.resize(ret.size()-k);
                tie(prev, cnt) = t.top(); t.pop();
            } 
        }        
        return ret;    
    }
};