LeetCode 솔루션

[5/23] 474. Ones and Zeroes

mingki 쪽지보내기 메일보내기 자기소개 아이디로 검색 전체게시물 작성
작성일 2022.05.24 08:38

6,983 조회
1 댓글
0 추천
0 비추천
목록

글검색

게시글 쓰기

[LeetCode 시즌 3] 2022년 5월 23일 문제입니다.

https://leetcode.com/problems/ones-and-zeroes/

474. Ones and Zeroes

Medium

3843383Add to ListShare

You are given an array of binary strings strs and two integers m and n.

Return the size of the largest subset of strs such that there are at most m 0's and n 1's in the subset.

A set x is a subset of a set y if all elements of x are also elements of y.

Example 1:

Input: strs = ["10","0001","111001","1","0"], m = 5, n = 3
Output: 4
Explanation: The largest subset with at most 5 0's and 3 1's is {"10", "0001", "1", "0"}, so the answer is 4.
Other valid but smaller subsets include {"0001", "1"} and {"10", "1", "0"}.
{"111001"} is an invalid subset because it contains 4 1's, greater than the maximum of 3.

Example 2:

Input: strs = ["10","0","1"], m = 1, n = 1
Output: 2
Explanation: The largest subset is {"0", "1"}, so the answer is 2.

Constraints:

1 <= strs.length <= 600
1 <= strs[i].length <= 100
strs[i] consists only of digits '0' and '1'.
1 <= m, n <= 100

#LeetCode, #Adobe, #Apple, #Amazon, #Google, #Array, #String, #Dynamic Programming(DP)

링크

https://leetcode.com/problems/ones-and-zeroes/ 1481 회 연결

익명
작성일 2022.05.24 12:13

knapsack problem
https://en.wikipedia.org/wiki/Knapsack_problem

C++
Runtime: 411 ms, faster than 45.44% of C++ online submissions for Ones and Zeroes.
Memory Usage: 9.8 MB, less than 82.68% of C++ online submissions for Ones and Zeroes.

class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        
        for (auto &s : strs) {
            vector<int> count(2, 0);
            for (auto c : s) count[c - '0']++;
            for (int zero = m; zero >= count[0]; --zero) {
                for (int one = n; one >= count[1]; --one) {
                    dp[zero][one] = max(1 + dp[zero - count[0]][one - count[1]], dp[zero][one]);
                }
            }
        }
        return dp[m][n];
    }
};