LeetCode 2397: Maximum Rows Covered by Columns
LeetCode 2397 Solution Explanation
Explanation:
To solve this problem, we can iterate over all possible combinations of selecting numSelect columns from the given matrix. For each combination, we check how many rows can be covered using those columns. We keep track of the maximum number of rows covered and return that as the result.
- Generate all combinations of selecting
numSelectcolumns. - For each combination, check how many rows can be covered.
- Return the maximum number of rows covered.
Time Complexity:
- Generating all combinations: O(2^n)
- Checking rows covered for each combination: O(m)
- Total time complexity: O(2^n * m)
Space Complexity:
The space complexity is O(1) as we are not using any additional data structures that grow with the input size.
:
LeetCode 2397 Solutions in Java, C++, Python
class Solution {
public int maxRowsCovered(int[][] matrix, int numSelect) {
int m = matrix.length;
int n = matrix[0].length;
int maxCovered = 0;
for (int comb = 0; comb < (1 << n); comb++) {
if (Integer.bitCount(comb) != numSelect) continue;
int rowsCovered = 0;
for (int row = 0; row < m; row++) {
boolean covered = true;
for (int col = 0; col < n; col++) {
if ((comb & (1 << col)) > 0 && matrix[row][col] == 1) {
covered = true;
break;
}
}
if (covered) rowsCovered++;
}
maxCovered = Math.max(maxCovered, rowsCovered);
}
return maxCovered;
}
}Interactive Code Editor for LeetCode 2397
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