How do you solve Matrix Diagonal Sum on LeetCode?

Matrix Diagonal Sum (LeetCode #1572) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The primary diagonal consists of elements where row index equals column index.

What is the brute force solution for Matrix Diagonal Sum?

The brute force approach for Matrix Diagonal Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves iterating through each element of the matrix to calculate the sums of both diagonals. This is straightforward but can be inefficient for larger matrices. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Matrix Diagonal Sum?

Matrix Diagonal Sum uses the following concepts: Array, Matrix. The recommended patterns to study are: Two Pointers, Sliding Window, Matrix Traversal.

What are the interview tips for Matrix Diagonal Sum?

Always clarify the problem constraints and edge cases. Discuss your thought process while coding to show your understanding. Consider both time and space complexities in your solution.

What are common mistakes when solving Matrix Diagonal Sum?

Counting the middle element twice in odd-sized matrices. Not considering the case when the matrix size is 1.

#1572

Matrix Diagonal Sum

Easy

ArrayMatrixTwo PointersSliding WindowMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution is similar to the brute force approach but focuses on reducing unnecessary operations. We can calculate both diagonal sums in a single loop, making it more efficient.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to hold the sum of the diagonals.
2Step 2: Loop through each row of the matrix, adding the primary and secondary diagonal elements in one pass.
3Step 3: If the matrix size is odd, subtract the middle element once.
4Step 4: Return the total sum.

solution.py9 lines

1def diagonalSum(mat):
2    n = len(mat)
3    total_sum = 0
4    for i in range(n):
5        total_sum += mat[i][i]  # Primary diagonal
6        total_sum += mat[i][n - 1 - i]  # Secondary diagonal
7    if n % 2 == 1:
8        total_sum -= mat[n // 2][n // 2]  # Subtract the middle element if odd
9    return total_sum

ℹ

Complexity note: The time complexity is O(n) because we only loop through the matrix once, and the space complexity is O(1) since we are using a constant amount of extra space.

1The primary diagonal consists of elements where row index equals column index.
2The secondary diagonal consists of elements where the column index is the reverse of the row index.

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