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".

Tips:

Time Complexity: O(kl + kmn), Space Complexity: O(mn)

DP, 公式为dp[i][j] = Math.max(1 + dp[i-count[0]][j-count[1]], dp[i][j])。

注意for (int i=m;i>=count[0];i--)，必须从后往前dp，因为从前往后的话会有重复使用字符串的问题。

You can try a simple example yourself, basically the difference is, if you loop in the forward way, then you may use a item more than once, for example Array = {"10", ...}, m = 5, n = 3, for the first element, you dp[1][1] = 1, then dp[2][2] = 2, dp[3][3] = 3. If the problem specifies you can use the strings more than once, then it would the algorithm to do it. Code:

public class Solution {
    public int findMaxForm(String[] strs, int m, int n) {
        int[][] dp = new int[m + 1][n + 1];
        for (String s : strs) {
            int[] count = count(s);
            for (int i = m; i >= count[0]; i--) {
                for (int j = n; j >= count[1]; j--) {
                    dp[i][j] = Math.max(dp[i][j], 1 + dp[i - count[0]][j - count[1]]);
                }
            }
        }
        return dp[m][n];
    }
    private int[] count(String s) {
        int[] res = new int[2];
        for (char c : s.toCharArray()) {
            if (c == '0') res[0]++;
            else res[1]++;
        }
        return res;
    }
}