0474. Ones and Zeroes (M)

题目

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

题意

给定一个包含仅由’0’和’1’组成的字符串的数组和整数m和n，要求选取最多的字符串，使其中的’0’个数不超过m，’1’个数不超过n。

思路

类似背包问题，使用动态规划解决。dp[i][j][k]表示在数组前i个中选出若干个字符串，使其中的’0’个数不超过j，’1’个数不超过k。对于第i个字符串，有两种情况，即选或者不选，记其中有zeros个’0’，ones个’1’，则可以得到递推式：

可以使用滚动数组进行空间优化。

代码实现

Java

动态规划

class Solution {
    public int findMaxForm(String[] strs, int m, int n) {
        int[][][] dp = new int[strs.length + 1][m + 1][n + 1];
        for (int i = 1; i <= strs.length; i++) {
            int zeros = 0, ones = 0;
            for (char c : strs[i - 1].toCharArray()) {
                if (c == '0') zeros++;
                else ones++;
            }
            for (int j = 0; j <= m; j++) {
                for (int k = 0; k <= n; k++) {
                    if (j - zeros < 0 || k - ones < 0) dp[i][j][k] = dp[i - 1][j][k];
                    else dp[i][j][k] = Math.max(dp[i - 1][j][k], dp[i - 1][j - zeros][k - ones] + 1);
                }
            }
        }
        return dp[strs.length][m][n];
    }
}

动态规划滚动数组优化

class Solution {
    public int findMaxForm(String[] strs, int m, int n) {
        int[][] dp = new int[m + 1][n + 1];
        for (String str : strs) {
            int zeros = 0, ones = 0;
            for (char c : str.toCharArray()) {
                if (c == '0') zeros++;
                else ones++;
            }
              // 必须先从大端开始，防止重复计算
            for (int i = m; i >= zeros; i--) {
                for (int j = n; j >= ones; j--) {
                    dp[i][j] = Math.max(dp[i][j], dp[i - zeros][j - ones] + 1);
                }
            }
        }
        return dp[m][n];
    }
}