# LC 474. Ones and Zeroes

### [Problem link](https://leetcode.com/problems/ones-and-zeroes/)

###### tags: `leedcode` `python` `c++` `medium` `DP`

You are given an array of binary strings <code>strs</code> and two integers <code>m</code> and <code>n</code>.

Return the size of the largest subset of <code>strs</code> such that there are **at most** <code>m</code> <code>0</code>'s and <code>n</code> <code>1</code>'s in the subset.

A set <code>x</code> is a **subset** of a set <code>y</code> if all elements of <code>x</code> are also elements of <code>y</code>.

**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:**

- <code>1 <= strs.length <= 600</code>
- <code>1 <= strs[i].length <= 100</code>
- <code>strs[i]</code> consists only of digits <code>'0'</code> and <code>'1'</code>.
- <code>1 <= m, n <= 100</code>

## Solution 1 - DP
#### Python

class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]

        for s in strs:
            ones = s.count('1')
            zeros = s.count('0')

            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[-1][-1]

#### C++

class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<vector<int>> dp(m + 1, vector<int>(n + 1 , 0));

        for (string &str: strs) {
            int cnt0 = 0;
            int cnt1 = 0;
            for (char &c: str) {
                if (c == '0') {
                    cnt0++;
                } else if (c == '1') {
                    cnt1++;
                }
            }
            for (int i = m; i >= cnt0; i--) {
                for (int j = n; j >= cnt1; j--) {
                    dp[i][j] = max(dp[i][j], dp[i - cnt0][j - cnt1] + 1);
                }
            }
        }
        return dp[m][n];
    }
};

>### Complexity
>k = strs.length
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(kmn) | O(mn) |

## Note
sol1:
先計算0與1的個數, 然後當作兩個維度的01背包.