# Ones And Zeroes Problem

## Description

LeetCode Problem 474.

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:

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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:

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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

## Sample C++ Code

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class Solution {
public:
int findMaxForm(vector<string>& strs, int m, int n) {
vector<vector<int>> memo(m+1, vector<int>(n+1, 0));
int numZeroes, numOnes;
for (auto &s : strs) {
numZeroes = numOnes = 0;
// count number of zeroes and ones in current string
for (auto c : s) {
if (c == '0')
numZeroes++;
else if (c == '1')
numOnes++;
}
// memo[i][j] = the max number of strings that can be formed with i 0's and j 1's
// from the first few strings up to the current string s
// Catch: have to go from bottom right to top left
// Why? If a cell in the memo is updated(because s is selected),
// we should be adding 1 to memo[i][j] from the previous iteration (when we were not considering s)
// If we go from top left to bottom right, we would be using results from this iteration => overcounting
for (int i = m; i >= numZeroes; i--) {
for (int j = n; j >= numOnes; j--) {
memo[i][j] = max(memo[i][j], memo[i - numZeroes][j - numOnes] + 1);
}
}
}
return memo[m][n];
}
};