Introduction

In this blog post, we will explore how to solve the LeetCode problem "474" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

• Dynamic Programming: Use a 3D DP table dp[i][j][k] to store the maximum subset size using the first i strings with at most j zeros and k ones.
  • Iterative Approach: Iterate through the strings, and for each string, decide whether to include it in the subset or not based on its counts of zeros and ones and the remaining capacity m and n.
  • Optimization: Build the DP table bottom-up, considering each string one by one to make inclusion/exclusion decisions.

• Runtime Complexity: O(l * m * n), where l is the length of strs. Storage Complexity: O(m * n)

Code

    def findMaxForm(strs, m, n):
    """
    Finds the size of the largest subset of strs with at most m 0's and n 1's.

    Args:
        strs: A list of binary strings.
        m: The maximum number of 0's allowed.
        n: The maximum number of 1's allowed.

    Returns:
        The size of the largest subset.
    """

    dp = [[0] * (n + 1) for _ in range(m + 1)]

    for s in strs:
        zeros = s.count('0')
        ones = s.count('1')
        for i in range(m, zeros - 1, -1):
            for j in range(n, ones - 1, -1):
                dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1)

    return dp[m][n]