474. Ones And Zeroes¶

Difficulty: Medium

474. Ones and Zeroes

Medium

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

Solution¶

class Solution {
    public int findMaxForm(String[] strs, int zero, int one) {
        int n = strs.length;
        int dp[][][] = new int[n + 1][zero + 1][one + 1];
        for (int current[][] : dp) for (int current1[] : current) Arrays.fill(current1, -1);
        int res = solve(0, strs, 0, 0, dp, zero, one);
        return res;
    }
    static int solve(int ind, String arr[] , int zero, int one, int dp[][][], int max_zero, int max_one) {
        if (ind >= arr.length) return 0;
        if (dp[ind][zero][one] != -1) return dp[ind][zero][one];
        int count0 = 0, count1 = 0, op1 = 0, op2 = 0;
        String current = arr[ind];
        for (int i = 0; i < current.length(); i++) {
            if (current.charAt(i) == '0') count0++;
            else count1++;
        }
        if (zero + count0 <= max_zero && one + count1 <= max_one) op1 = 1 + solve(ind + 1, arr, zero + count0 , one + count1 , dp, max_zero, max_one);
        op2 = solve(ind + 1, arr, zero, one, dp, max_zero, max_one);
        return dp[ind][zero][one] = Math.max(op1, op2);
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here