#2643

Easy

LeetCode Problem Workspace

Row With Maximum Ones

Find the row with the maximum ones in a binary matrix and return its index and count.

Problem Statement

You are given an m x n binary matrix, mat, where each element is either a 0 or 1. Your task is to identify the row that contains the highest count of ones.

If there are multiple rows with the same number of ones, return the row with the smallest index. Your solution should return an array consisting of the row index and the number of ones in that row.

Examples

Example 1

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

Output: [0,1]

Both rows have the same number of 1's. So we return the index of the smaller row, 0, and the maximum count of ones (1). So, the answer is [0,1].

Example 2

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

Output: [1,2]

The row indexed 1 has the maximum count of ones (2). So we return its index, 1, and the count. So, the answer is [1,2].

Example 3

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

Output: [1,2]

The row indexed 1 has the maximum count of ones (2). So the answer is [1,2].

Constraints

m == mat.length
n == mat[i].length
1 <= m, n <= 100
mat[i][j] is either 0 or 1.

Solution Approach

Iterate Through Rows

To find the row with the maximum number of ones, iterate through each row of the matrix. For each row, count the ones and track the maximum count along with the corresponding row index.

Efficient Counting

You can optimize counting by iterating over the row elements and updating the count of ones in each row, keeping track of the current row with the maximum ones.

Handle Multiple Rows with Same Count

In cases where multiple rows have the same count of ones, ensure you return the row with the smallest index, as this is specified in the problem constraints.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the final approach but generally involves iterating through the rows, making the complexity O(m * n), where m is the number of rows and n is the number of columns. Space complexity is O(1) if only the count and index are stored during iteration.

What Interviewers Usually Probe

Tests the candidate's ability to work with matrix traversal and counting techniques.
Checks if the candidate can identify the smallest index in the case of multiple maximum counts.
Assesses how well the candidate manages efficiency for larger matrices.

Common Pitfalls or Variants

Common pitfalls

Not handling cases with multiple rows having the same number of ones correctly.
Failing to consider the matrix's size and making the solution inefficient for larger inputs.
Incorrectly implementing the count of ones or missing edge cases, such as an empty row or no ones in a row.

Follow-up variants

What if the matrix has only zeros? The algorithm should still return a valid row, even if all counts are zero.
What if the matrix contains only one row or one column? Consider edge cases where the matrix is minimal.
If we wanted to find the row with the least number of ones instead, how would that change the solution?

FAQ

How do I approach the Row With Maximum Ones problem?

Start by iterating through each row of the matrix and counting the ones in each row. Track the row with the highest count, and if counts are equal, return the smallest indexed row.

What is the time complexity of the solution?

The time complexity is O(m * n), where m is the number of rows and n is the number of columns, due to the need to iterate over each element in the matrix.

How do I handle multiple rows with the same count of ones?

Simply track the smallest index of the row with the maximum number of ones. If two rows have the same count, return the row with the smaller index.

What if the matrix is empty or contains only zeros?

Even in these cases, the algorithm should return the first row with zero ones. Handle edge cases like these to ensure robustness.

How can I improve the efficiency of the algorithm?

Ensure that you stop iterating over rows early if possible, and avoid unnecessary recalculations. Using a more advanced approach like binary search might speed up counting ones in sorted rows.

terminal

Solution

Solution 1: Simulation

We initialize an array $\textit{ans} = [0, 0]$ to store the index of the row with the most $1$s and the count of $1$s.

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class Solution:
    def rowAndMaximumOnes(self, mat: List[List[int]]) -> List[int]:
        ans = [0, 0]
        for i, row in enumerate(mat):
            cnt = sum(row)
            if ans[1] < cnt:
                ans = [i, cnt]
        return ans

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