LeetCode 474: Ones and Zeroes Solution

Master LeetCode problem 474 (Ones and Zeroes), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

474. Ones and Zeroes

Medium

Array String Dynamic Programming

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming to build a 3D array dp where dp[i][j][k] represents the maximum size of subset we can form using the first i strings with at most j zeros and k ones. We iterate through each string in the input array and update the dp array accordingly.

Algorithmic Idea:

Initialize a 3D array dp with dimensions (strs.length + 1) x (m + 1) x (n + 1) and initialize all values to 0.
For each string in strs, count the number of zeros and ones.
Iterate through the dp array and update the values based on whether we include the current string or not.
Finally, return the value at dp[strs.length][m][n].

Time Complexity:

The time complexity of this solution is O(l * m * n) where l is the number of strings in the input array.

Space Complexity:

The space complexity of this solution is O(l * m * n) where l is the number of strings in the input array.

Solution Code

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

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