Leetcode 474. 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 0s and 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".

Analysis:
This is a classic backpack DP problem. Use dp[m + 1][n + 1], where dp[i][j] denotes for number of i 1s and j 0s, the maximum number of strings. 

Code (Java):

class Solution {
    public int findMaxForm(String[] strs, int m, int n) {
        if (strs == null || strs.length == 0) {
            return 0;
        }
        
        int[][] dp = new int[m + 1][n + 1];
        
        for (int i = 1; i <= strs.length; i++) {
            for (int j = m; j >= 0; j--) {
                for (int k = n; k >= 0; k--) {
                    int num0 = 0;
                    int num1 = 0;
                    int[] count = countNumberOfZerosAndOnes(strs[i - 1]);
                    num0 = count[0];
                    num1 = count[1];
                    
                    // no pick
                    //
                    int noPick = dp[j][k];
                    
                    // pick
                    //
                    int pick = 0;
                    if (j - num0 >= 0 && k - num1 >= 0) {
                        pick = dp[j - num0][k - num1] + 1;    
                    }
                    
                    dp[j][k] = Math.max(pick, noPick);
                }
            }
        }
        
        return dp[m][n];
    }
    
    private int[] countNumberOfZerosAndOnes(String s) {
        int zero = 0;
        int one = 0;
        
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            
            if (c == '0') {
                zero++;
            } else {
                one++;
            }
        }
        
        return new int[]{zero, one};
    }
}