LeetCode 474. Ones and Zeroes

原题网址：https://leetcode.com/problems/ones-and-zeroes/

In the computer world, use restricted resource you have to generate maximum benefit is what we always want to pursue.

For now, suppose you are a dominator of m 0s and n 1s respectively. On the other hand, there is an array with strings consisting of only 0sand 1s.

Now your task is to find the maximum number of strings that you can form with given m 0s and n 1s. Each 0 and 1 can be used at most once.

Note:

1. The given numbers of 0s and 1s will both not exceed 100
2. The size of given string array won't exceed 600.

Example 1:

Input: Array = {"10", "0001", "111001", "1", "0"}, m = 5, n = 3
Output: 4

Explanation: This are totally 4 strings can be formed by the using of 5 0s and 3 1s, which are “10,”0001”,”1”,”0”

Example 2:

Input: Array = {"10", "0", "1"}, m = 1, n = 1
Output: 2

Explanation: You could form "10", but then you'd have nothing left. Better form "0" and "1".

方法：动态规划。

Java代码：

public class Solution {
    public int findMaxForm(String[] strs, int m, int n) {
        int[][] max = new int[m+1][n+1];
        for(String str : strs) {
            int[] counts = count(str);
            for(int i = m; i >= counts[0]; i--) {
                for(int j = n; j >= counts[1]; j--) {
                    max[i][j] = Math.max(max[i][j], 1 + max[i - counts[0]][j - counts[1]]);
                }
            }
        }
        return max[m][n];
        
    }
    private int[] count(String str) {
        char[] sa = str.toCharArray();
        int[] counts = new int[2];
        for(int i = 0; i < sa.length; i++) {
            counts[sa[i] - '0'] ++;
        }
        return counts;
    }
}

Python代码：

class Solution(object):
    def findMaxForm(self, strs, m, n):
        """
        :type strs: List[str]
        :type m: int
        :type n: int
        :rtype: int
        """
        dp = [[0 for x in range(n + 1)] for y in range(m + 1)]
        for str in strs:
            c0, c1 = str.count("0"), str.count("1")
            for i in range(m, c0 - 1, -1):
                for j in range(n, c1 - 1, -1):
                    dp[i][j] = max(dp[i][j], 1 + dp[i - c0][j - c1])
        return dp[m][n]