Row With Maximum Ones - Practice Coding Problems

Row With Maximum Ones - Problem

Array Matrix Easy

You are analyzing a binary matrix where each cell contains either 0 or 1. Your task is to identify which row contains the maximum number of ones and return both the row index and the count of ones in that row.

Given an m × n binary matrix mat, find the 0-indexed position of the row that contains the maximum count of ones. If multiple rows tie for the maximum count, choose the row with the smallest index.

Goal: Return an array [rowIndex, onesCount] where rowIndex is the index of the row with maximum ones, and onesCount is the number of ones in that row.

Example: In matrix [[0,1],[1,0]], both rows have 1 one, so we return [0, 1] (row 0 wins the tie).

Input & Output

example_1.py — Basic Matrix

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

› Output: [0,1]

💡 Note: Both rows have exactly 1 one. Since row 0 comes first (smaller index), we return [0,1].

example_2.py — Clear Winner

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

› Output: [1,2]

💡 Note: Row 0 has 0 ones, Row 1 has 2 ones. Row 1 has the maximum, so we return [1,2].

example_3.py — All Zeros

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

› Output: [0,0]

💡 Note: All rows have 0 ones. Row 0 wins the tie (smallest index), so we return [0,0].

Constraints

m == mat.length
n == mat[i].length
1 ≤ m, n ≤ 100
mat[i][j] is either 0 or 1
Matrix contains only binary values (0 or 1)

Visualization

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Understanding the Visualization

Survey Each Section

Walk through each row (section) and count occupied seats (1s)

Track the Winner

Keep track of which section has the most people so far

Handle Ties

If sections tie, choose the one with the lower number (earlier in the stadium)

Report Results

Return both the section number and how many people are in it

Key Takeaway

🎯 Key Insight: We must examine every row to find the maximum, but we can efficiently track the best result in a single pass through the matrix.

Asked in

G Google 15 a Amazon 12 ⊞ Microsoft 8 f Meta 6

The optimal approach uses a single pass through the matrix to count ones in each row while tracking the maximum. This achieves O(m×n) time complexity which is optimal since we must examine every element. The key insight is that we need to check every row to find the true maximum, but we can do this efficiently by maintaining the best result as we scan.

Common Approaches

Approach	Time	Space	Notes
✓ Single Pass Row Counting	O(m × n)	O(1)	Iterate through each row, count ones, and track the maximum

Single Pass Row Counting — Algorithm Steps

Initialize variables to track the best row index and maximum ones count
Iterate through each row of the matrix
For each row, count the number of ones
If current row's count exceeds our maximum, update the result
Return the row index and count of the best row found

Visualization

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Step-by-Step Walkthrough

Start with first row

Begin scanning from row 0, count=0

Count ones in current row

Sum up all 1s in the current row

Update maximum if needed

If current count > max, update best row and count

Move to next row

Continue until all rows are processed

Code -

solution.c — C

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

int* rowAndMaximumOnes(int** mat, int matSize, int* matColSize, int* returnSize) {
    int maxCount = -1;
    int maxRow = 0;
    
    for (int i = 0; i < matSize; i++) {
        int count = 0;
        for (int j = 0; j < matColSize[i]; j++) {
            count += mat[i][j];
        }
        if (count > maxCount) {
            maxCount = count;
            maxRow = i;
        }
    }
    
    int* result = (int*)malloc(2 * sizeof(int));
    result[0] = maxRow;
    result[1] = maxCount;
    *returnSize = 2;
    return result;
}

int main() {
    // Sample usage
    int row1[] = {0, 1};
    int row2[] = {1, 0};
    int* mat[] = {row1, row2};
    int colSizes[] = {2, 2};
    int returnSize;
    int* result = rowAndMaximumOnes(mat, 2, colSizes, &returnSize);
    printf("[%d,%d]\n", result[0], result[1]);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m × n)

We need to examine every element in the m×n matrix to count ones in each row

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to track the maximum count and row index

✓ Linear Space

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Medium Frequency

~8 min Avg. Time

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Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Single Pass Row Counting — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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