LC474.md

Problem: Ones and Zeros

Difficulty: Medium

Description:

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.

Examples:

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.

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

Solutions:

For the naive approach, we can use the stack data structure to track the subsets that meet the conditions. This approach has a time complexity of O(N^3) and space complexity of O(N^2).

def findMaxForm(strs, m, n):
    """
    :type strs: List[str]
    :type m: int
    :type n: int
    :rtype: int
    :Time Comeplexity: O(N^3)
    :Space Complexity: O(N^2)
    """
    ans = 0
    for i in range(len(strs)):
        stack = []
        stack.append(strs[i])
        for j in range(len(strs)):
            if i != j:
                stack.append(strs[j])
                tmp = "".join(stack)
                if tmp.count("1") <= n and tmp.count("0") <= m and len(tmp) <= n + m:
                    ans = max(len(stack), ans)
                else:
                    stack.pop()

    return ans

For the DP approach we create an m by n array to track the maximum number of elements for each possible of m and n. This solution has a time complexity of O(N * m * n) while it does have a space complexity of O(m * n).

def findMaxForm(strs, m, n):
    """
    :type strs: List[str]
    :type m: int
    :type n: int
    :rtype: int
    :Time Comeplexity: O(N * m * n)
    :Space Complexity: O(m * n)
    """
    dp = [[0 for _ in range(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 - zeros][j - ones] + 1, dp[i][j])

    return dp[-1][-1]