## DEV Community

Abhishek Chaudhary

Posted on

# 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)
``````