LeetCode #2397 — MEDIUM

Maximum Rows Covered by Columns

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an m x n binary matrix matrix and an integer numSelect.

Your goal is to select exactly numSelect distinct columns from matrix such that you cover as many rows as possible.

A row is considered covered if all the 1's in that row are also part of a column that you have selected. If a row does not have any 1s, it is also considered covered.

More formally, let us consider selected = {c₁, c₂, ...., c_numSelect} as the set of columns selected by you. A row i is covered by selected if:

For each cell where matrix[i][j] == 1, the column j is in selected.
Or, no cell in row i has a value of 1.

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

Example 1:

Input: matrix = [[0,0,0],[1,0,1],[0,1,1],[0,0,1]], numSelect = 2

Output: 3

Explanation:

One possible way to cover 3 rows is shown in the diagram above.
We choose s = {0, 2}.
- Row 0 is covered because it has no occurrences of 1.
- Row 1 is covered because the columns with value 1, i.e. 0 and 2 are present in s.
- Row 2 is not covered because matrix[2][1] == 1 but 1 is not present in s.
- Row 3 is covered because matrix[2][2] == 1 and 2 is present in s.
Thus, we can cover three rows.
Note that s = {1, 2} will also cover 3 rows, but it can be shown that no more than three rows can be covered.

Example 2:

Input: matrix = [[1],[0]], numSelect = 1

Output: 2

Explanation:

Selecting the only column will result in both rows being covered since the entire matrix is selected.

Constraints:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 12
matrix[i][j] is either 0 or 1.
1 <= numSelect <= n

Step 01

Brute Force Baseline

Problem summary: You are given an m x n binary matrix matrix and an integer numSelect. Your goal is to select exactly numSelect distinct columns from matrix such that you cover as many rows as possible. A row is considered covered if all the 1's in that row are also part of a column that you have selected. If a row does not have any 1s, it is also considered covered. More formally, let us consider selected = {c1, c2, ...., cnumSelect} as the set of columns selected by you. A row i is covered by selected if: For each cell where matrix[i][j] == 1, the column j is in selected. Or, no cell in row i has a value of 1. Return the maximum number of rows that can be covered by a set of numSelect columns.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Backtracking · Bit Manipulation

Example 1

[[0,0,0],[1,0,1],[0,1,1],[0,0,1]]
2

Example 2

[[1],[0]]
1

Core Insight

What unlocks the optimal approach

Try a brute-force approach.
Iterate through all possible sets of exactly <code>cols</code> columns.
For each valid set, check how many rows are covered, and return the maximum.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2397: Maximum Rows Covered by Columns
class Solution {
    public int maximumRows(int[][] matrix, int numSelect) {
        int m = matrix.length, n = matrix[0].length;
        int[] rows = new int[m];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] == 1) {
                    rows[i] |= 1 << j;
                }
            }
        }
        int ans = 0;
        for (int mask = 1; mask < 1 << n; ++mask) {
            if (Integer.bitCount(mask) != numSelect) {
                continue;
            }
            int t = 0;
            for (int x : rows) {
                if ((x & mask) == x) {
                    ++t;
                }
            }
            ans = Math.max(ans, t);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2397: Maximum Rows Covered by Columns
func maximumRows(matrix [][]int, numSelect int) (ans int) {
	m, n := len(matrix), len(matrix[0])
	rows := make([]int, m)
	for i, row := range matrix {
		for j, x := range row {
			if x == 1 {
				rows[i] |= 1 << j
			}
		}
	}
	for mask := 1; mask < 1<<n; mask++ {
		if bits.OnesCount(uint(mask)) != numSelect {
			continue
		}
		t := 0
		for _, x := range rows {
			if (x & mask) == x {
				t++
			}
		}
		if ans < t {
			ans = t
		}
	}
	return
}

# Accepted solution for LeetCode #2397: Maximum Rows Covered by Columns
class Solution:
    def maximumRows(self, matrix: List[List[int]], numSelect: int) -> int:
        rows = []
        for row in matrix:
            mask = reduce(or_, (1 << j for j, x in enumerate(row) if x), 0)
            rows.append(mask)

        ans = 0
        for mask in range(1 << len(matrix[0])):
            if mask.bit_count() != numSelect:
                continue
            t = sum((x & mask) == x for x in rows)
            ans = max(ans, t)
        return ans

// Accepted solution for LeetCode #2397: Maximum Rows Covered by Columns
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2397: Maximum Rows Covered by Columns
// class Solution {
//     public int maximumRows(int[][] matrix, int numSelect) {
//         int m = matrix.length, n = matrix[0].length;
//         int[] rows = new int[m];
//         for (int i = 0; i < m; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 if (matrix[i][j] == 1) {
//                     rows[i] |= 1 << j;
//                 }
//             }
//         }
//         int ans = 0;
//         for (int mask = 1; mask < 1 << n; ++mask) {
//             if (Integer.bitCount(mask) != numSelect) {
//                 continue;
//             }
//             int t = 0;
//             for (int x : rows) {
//                 if ((x & mask) == x) {
//                     ++t;
//                 }
//             }
//             ans = Math.max(ans, t);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2397: Maximum Rows Covered by Columns
function maximumRows(matrix: number[][], numSelect: number): number {
    const [m, n] = [matrix.length, matrix[0].length];
    const rows: number[] = Array(m).fill(0);
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (matrix[i][j]) {
                rows[i] |= 1 << j;
            }
        }
    }
    let ans = 0;
    for (let mask = 1; mask < 1 << n; ++mask) {
        if (bitCount(mask) !== numSelect) {
            continue;
        }
        let t = 0;
        for (const x of rows) {
            if ((x & mask) === x) {
                ++t;
            }
        }
        ans = Math.max(ans, t);
    }
    return ans;
}

function bitCount(i: number): number {
    i = i - ((i >>> 1) & 0x55555555);
    i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
    i = (i + (i >>> 4)) & 0x0f0f0f0f;
    i = i + (i >>> 8);
    i = i + (i >>> 16);
    return i & 0x3f;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(2^n × m)

Space

O(m)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.