Row With Maximum Ones

Row With Maximum Ones - Problem

Array Matrix Easy

Given an m x n binary matrix mat, find the 0-indexed position of the row that contains the maximum count of ones, and the number of ones in that row.

In case there are multiple rows that have the maximum count of ones, the row with the smallest row number should be selected.

Return an array containing the index of the row, and the number of ones in it.

Input & Output

Example 1 — Basic Case

$ Input: mat = [[0,1,1,1],[0,0,1,1],[0,0,0,1]]

› Output: [0,3]

💡 Note: Row 0 has 3 ones, Row 1 has 2 ones, Row 2 has 1 one. Maximum is 3 in row 0.

Example 2 — Tie Breaking

$ Input: mat = [[0,0],[1,1],[0,0]]

› Output: [1,2]

💡 Note: Only Row 1 has ones (2 ones), so return [1,2].

Example 3 — Multiple Ties

$ Input: mat = [[0,1],[1,0],[0,1]]

› Output: [0,1]

💡 Note: Rows 0, 1, and 2 all have 1 one each. Return smallest row index [0,1].

Constraints

1 ≤ m, n ≤ 100
mat[i][j] is either 0 or 1

Visualization

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Asked in

M Microsoft 15 a Amazon 12

The key insight is to count ones in each row sequentially, which naturally handles tie-breaking by preferring earlier rows. Best approach is simple iteration with optional early exit optimization. Time: O(m×n), Space: O(1).

Common Approaches

✓ Brute Force - Count Ones in Each Row

⏱️ Time: O(m×n) Space: O(1)

Iterate through each row, count the number of ones, and keep track of the row index with the maximum count. Since we process rows from top to bottom, ties are automatically resolved in favor of the earlier row.

Optimized - Early Exit on All Ones

⏱️ Time: O(m×n) Space: O(1)

Count ones in each row sequentially, but if we find a row with all ones (equal to number of columns), we can immediately return since no row can have more ones than that.

Brute Force - Count Ones in Each Row — Algorithm Steps

Step 1: Initialize variables to track max count and best row index
Step 2: For each row, count the number of ones
Step 3: Update best row if current count is greater than max count
Step 4: Return the best row index and its count

Visualization

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Step-by-Step Walkthrough

Initialize

Set maxCount = -1, maxRow = 0

Count Each Row

Count ones in each row sequentially

Update Best

Update maxRow when count > maxCount

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void solution(int mat[][100], int rows, int cols, int* result) {
    int maxCount = -1;
    int maxRow = 0;
    
    for (int i = 0; i < rows; i++) {
        int count = 0;
        for (int j = 0; j < cols; j++) {
            count += mat[i][j];
        }
        if (count > maxCount) {
            maxCount = count;
            maxRow = i;
        }
    }
    
    result[0] = maxRow;
    result[1] = maxCount;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parser for 2D array
    int mat[100][100];
    int rows = 0, cols = 0;
    
    char* ptr = line + 1; // Skip first '['
    while (*ptr && *ptr != ']') {
        if (*ptr == '[') {
            ptr++;
            cols = 0;
            while (*ptr && *ptr != ']') {
                if (*ptr >= '0' && *ptr <= '1') {
                    mat[rows][cols++] = *ptr - '0';
                }
                ptr++;
            }
            rows++;
        }
        ptr++;
    }
    
    int result[2];
    solution(mat, rows, cols, result);
    printf("[%d,%d]\n", result[0], result[1]);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m×n)

We visit each cell in the m×n matrix exactly once to count ones

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to track max count and row index

✓ Linear Space

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Medium Frequency

~8 min Avg. Time

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Ln 1, Col 1

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Count Ones in Each Row — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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