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						[5/23] 474. Ones and Zeroes
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[LeetCode 시즌 3] 2022년 5월 23일 문제입니다.
https://leetcode.com/problems/ones-and-zeroes/
474. Ones and Zeroes
Medium
3843383Add to ListShareYou 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
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					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.
				
													
								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];
    }
}; 
								 
							







