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Ones and Zeroes⚓︎

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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.

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⚓︎

Way 1⚓︎

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class Solution {
private:
    int count(const string &str, const char c) const {
        int cnt = 0;
        for (char ch : str) {
            if (ch == c) cnt++;
        }
        return cnt;
    }
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<int> zeros(strs.size() + 1), ones(strs.size() + 1);
        for (int i = 1; i <= strs.size(); i++) {
            zeros[i] = count(strs[i - 1], '0');
            ones[i] = count(strs[i - 1], '1');
        }
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        dp[0][0] = 0;

        for (int i = 1; i <= strs.size(); i++) {
            for (int j = m; j >= zeros[i]; j--) {
                for (int k = n; k >= ones[i]; k--) {
                    dp[j][k] = max(dp[j][k], dp[j - zeros[i]][k - ones[i]] + 1);
                }
            }
        }

        return dp[m][n];
    }
};

Another way of writing:

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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 one = 0, zero = 0;
            for (char c : str) {
                (c == '0') ? zero++ : one++;
            }
            for (int i = m; i >= zero; i--) {
                for (int j = n; j >= one; j--) {
                    dp[i][j] = max(dp[i][j], dp[i - zero][j - one] + 1);
                }
            }
        }

        return dp[m][n];
    }
};
  • Time complexity: \(O(knm)\), where \(k\) is the number of elements of strs;
  • Space complexity: \(O(mn)\).