Maximum Rows Covered by Columns LeetCode Solution

Problem – Maximum Rows Covered by Columns LeetCode Solution

You are given a 0-indexed m x n binary matrix mat and an integer cols, which denotes the number of columns you must choose.

A row is covered by a set of columns if each cell in the row that has a value of 1 also lies in one of the columns of the chosen set.

Return the maximum number of rows that can be covered by a set of cols columns.

Example 1:

Input: mat = [[0,0,0],[1,0,1],[0,1,1],[0,0,1]], cols = 2
Output: 3
Explanation:
As shown in the diagram above, one possible way of covering 3 rows is by selecting the 0th and 2nd columns.
It can be shown that no more than 3 rows can be covered, so we return 3.

Example 2:

Input: mat = [[1],[0]], cols = 1
Output: 2
Explanation:
Selecting the only column will result in both rows being covered, since the entire matrix is selected.
Therefore, we return 2.

Constraints:

  • m == mat.length
  • n == mat[i].length
  • 1 <= m, n <= 12
  • mat[i][j] is either 0 or 1.
  • 1 <= cols <= n

Maximum Rows Covered by Columns LeetCode Solution in C++

class Solution {
public:
    int maxi = INT_MIN;
    void helper(vector<vector<int>> &mat , int m , int n , int cols , int idx , vector<int> &vis){
        if(cols == 0 or idx==n){
            int cnt = 0;
            for(int p = 0 ; p < m ; p++){
                bool check = true;
                for(int q = 0 ; q < n; q++){
                    // if cell is 1 and not visited then we cannot take this row
                    if(mat[p][q] == 1 and vis[q] == 0){
                        check = false;
                        break;
                    }
                }
                if(check) cnt++;
            }
            maxi = max(maxi,cnt);
            return;
        }
        
        // picking idx th column and marking column as visited
        vis[idx]=1;
        helper(mat,m,n,cols-1,idx+1,vis);
        vis[idx]=0;
        
        // not picking
        helper(mat,m,n,cols,idx+1,vis);
        return;
        
    }
    int maximumRows(vector<vector<int>>& mat, int cols) {
        int m = mat.size();
        int n = mat[0].size();
        
        vector<int> vis(n);
        
        helper(mat,m,n,cols,0,vis);
        return maxi;
    }
};

Maximum Rows Covered by Columns LeetCode Solution in Java

    static int maximumRows(int[][] mat, int cols) {
        return getResult(mat, cols, new HashSet<Integer>(),0) ;
    }
    static int getResult(int [][] mat, int cols, Set<Integer> set, int x){
        int maxResult = 0;
        if(cols == 0){
            int count = 0;
            for(int i = 0; i < mat.length; i++) {
                int [] row = mat[i];
                boolean flag = true;
                for(int col = 0; col < row.length; col++){
                    if(row[col] == 1 && !set.contains(col)){
                        flag = false; break;
                    }
                }
                if(flag) count++;

            }
            return count;
        }

        for(int i = x; i < mat[0].length; i++) {
            set.add(i);
            maxResult = Math.max(maxResult, getResult(mat, cols - 1, set, i+1));
            set.remove(i);
        }
        return maxResult;
    }

Maximum Rows Covered by Columns LeetCode Solution in Python

class Solution:
    def maximumRows(self, mat: List[List[int]], cols: int) -> int:
        n,m = len(mat),len(mat[0])
        ans = 0

        def check(state,row,rowIncludedCount):
            nonlocal ans
            if row==n:
                if sum(state)<=cols:
                    ans = max(ans,rowIncludedCount)
                return
            
            check(state[::],row+1,rowIncludedCount)
            for j in range(m):
                if mat[row][j]==1:
                    state[j] = 1
            check(state,row+1,rowIncludedCount+1)
        
        check([0]*m,0,0)
        return ans
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