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Abhishek Chaudhary
Abhishek Chaudhary

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Ones and Zeroes

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:
    def longest(self, strs, i, m, n, slen):
        if m < 0 or n < 0:
            return -1
        if i >= slen:
            return 0
        key = (i, m, n)
        if key in self.cache:
            return self.cache[key]
        x, y = strs[i]
        a = 1 + self.longest(strs, i + 1, m - x, n - y, slen)
        b = self.longest(strs, i + 1, m, n, slen)
        mlen = max(a, b)
        self.cache[key] = mlen
        return mlen

    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        slen = len(strs)
        strs = [(s.count("0"), s.count("1")) for s in strs]
        self.cache = {}
        return self.longest(strs, 0, m, n, slen)
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