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Leetcode Ones and Zeroes problem solution

YASH PAL, 31 July 2024

In this Leetcode Ones and Zeroes problem solution 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.

Leetcode Ones and Zeroes problem solution

Problem solution in Python.

from collections import Counter
class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        weights = []
        for st in strs:
            counter = Counter(st)
            weights.append((counter.get("0", 0), counter.get("1", 0)))
        self.visited = {}
        return self.recursiveWithMemo(strs, weights, len(strs) - 1, (m, n))
    
    # Brute Force Recursive Solution
    def recursiveWithoutMemo(self, strs, weights, index, wt):
        if wt == (0, 0) or index < 0: return 0
        else:
            curr = weights[index]
            if curr[0] <= wt[0] and curr[1] <= wt[1]:
                return max(
                    1 + self.recursiveWithoutMemo(strs, weights, index - 1, (wt[0] - curr[0], wt[1] - curr[1])),
                    self.recursiveWithoutMemo(strs, weights, index - 1, wt)
                )
            else:
                return self.recursiveWithoutMemo(strs, weights, index - 1, wt)

Problem solution in Java.

public int findMaxForm(String[] strs, int m, int n) {
        Arrays.sort(strs, new Comparator<String>(){
            public int compare(String s1, String s2) {
                int val = s1.length() - s2.length();
                if(val != 0)
                    return val;
                return s1.compareTo(s2);
            }
        });
        int result = 0;
        for(int i = 0; i < strs.length; i++)
            result = Math.max(result, dfs(strs, m, n, i));
        return result;
    }
    public int dfs(String[] strs, int m, int n, int start) {
        boolean flag;
        int result = 0, mm, nn;
        for(; start < strs.length; start++) {
            char[] c = strs[start].toCharArray();
            flag = true;
            if(c.length > m + n)
                break;
            mm = m;
            nn = n;
            for(int i = 0; i < c.length; i++) {
                if(c[i] == '0' && m > 0)
                    m--;
                else if(c[i] == '1' && n > 0)
                    n--;
                else {
                    flag = false;
                    break;
                }
            }
            if(flag)
                result++;
            else {
                m = mm;
                n = nn;
            }
        }
        return result;
    }

Problem solution in C++.

pair<int, int> countChars(string& str) {
    int ones = 0, zeros = 0;
    for (auto c : str) {
        if (c == '0') {
            zeros++;
        } else {
            ones++;
        }
    }
    return {zeros, ones};
}
int findMaxForm(vector<string>& strs, int m, int n) {
    vector<vector<vector<int>>> dp(strs.size() + 1, vector<vector<int>>(m + 1, vector<int>(n + 1, 0)));
    for (int i = 0; i < strs.size(); i++) {
        auto counts = countChars(strs[i]);
        int ones = counts.second, zeros = counts.first;
        for (int j = 0; j <= m; j++) {
            for (int k = 0; k <= n; k++) {
                if (zeros <= j && ones <= k) {
                    dp[i + 1][j][k] = max(dp[i][j][k], 1 + dp[i][j - zeros][k - ones]);
                } else {
                    dp[i + 1][j][k] = dp[i][j][k];
                }
            }
        }
    }
    return dp[strs.size()][m][n];
}

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