How do you solve Row With Maximum Ones on LeetCode?

Row With Maximum Ones (LeetCode #2643) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m + n) time complexity and O(1) space complexity. Key insight: Using a two-pointer technique can significantly reduce the number of checks needed.

What is the brute force solution for Row With Maximum Ones?

The brute force approach for Row With Maximum Ones is Brute Force, with O(m * n) time complexity and O(1) space complexity. The brute force approach involves checking each row one by one to count the number of ones. This is straightforward but can be slow for larger matrices since we check every element. The optimal Optimal Solution approach improves this to O(m + n).

What algorithm or data structure is used to solve Row With Maximum Ones?

Row With Maximum Ones uses the following concepts: Array, Matrix. The recommended patterns to study are: Two Pointers, Array.

What are the interview tips for Row With Maximum Ones?

Always clarify the problem requirements and constraints before jumping into coding. Think about how you can optimize your solution after implementing the brute force approach. Practice explaining your thought process as you code; it helps during interviews.

What are common mistakes when solving Row With Maximum Ones?

Not considering edge cases where multiple rows have the same maximum count. Failing to optimize the counting process, leading to unnecessary computations.

#2643

Row With Maximum Ones

Easy

ArrayMatrixTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n)	O(m + n)
Space	O(1)	O(1)

💡

Intuition

Time O(m + n)Space O(1)

The optimal solution leverages the fact that the matrix is binary. By using a two-pointer technique, we can efficiently find the row with the maximum ones without counting each row from scratch.

⚙️

Algorithm

5 steps

1Step 1: Initialize two pointers, one at the start of the first row and the other at the end of the last column.
2Step 2: Traverse the matrix using the two pointers. If the current cell is 1, move the row pointer down and update the count of ones.
3Step 3: If the current cell is 0, move the column pointer left to check for more ones.
4Step 4: Keep track of the maximum count of ones and the corresponding row index.
5Step 5: Return the row index and the maximum count of ones.

solution.py15 lines

1def rowWithMaximumOnes(mat):
2    max_count = 0
3    row_index = 0
4    n = len(mat[0])
5    row, col = 0, n - 1
6    while row < len(mat) and col >= 0:
7        if mat[row][col] == 1:
8            count = n - col
9            if count > max_count:
10                max_count = count
11                row_index = row
12            row += 1
13        else:
14            col -= 1
15    return [row_index, max_count]

ℹ

Complexity note: We traverse the matrix using two pointers, which allows us to skip unnecessary checks, leading to a time complexity of O(m + n). Space complexity remains O(1) as we only use a few variables.

1Using a two-pointer technique can significantly reduce the number of checks needed.
2Understanding the structure of the matrix (binary) allows for optimizations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.