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Formatted question description: https://leetcode.ca/all/1572.html

1572. Matrix Diagonal Sum (Easy)

Given a square matrix mat, return the sum of the matrix diagonals.

Only include the sum of all the elements on the primary diagonal and all the elements on the secondary diagonal that are not part of the primary diagonal.

 

Example 1:

Input: mat = [[1,2,3],
              [4,5,6],
              [7,8,9]]
Output: 25
Explanation: Diagonals sum: 1 + 5 + 9 + 3 + 7 = 25
Notice that element mat[1][1] = 5 is counted only once.

Example 2:

Input: mat = [[1,1,1,1],
              [1,1,1,1],
              [1,1,1,1],
              [1,1,1,1]]
Output: 8

Example 3:

Input: mat = [[5]]
Output: 5

 

Constraints:

  • n == mat.length == mat[i].length
  • 1 <= n <= 100
  • 1 <= mat[i][j] <= 100

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/matrix-diagonal-sum/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int diagonalSum(vector<vector<int>>& A) {
        int N = A.size(), sum = 0;
        for (int i = 0; i < N; ++i) sum += i == N - i - 1 ? A[i][i] : (A[i][i] + A[N - i - 1][i]);
        return sum;
    }
};

Java

  • class Solution {
    public int diagonalSum(int[][] mat) {
        int sum = 0;
        int side = mat.length;
        for (int i = 0; i < side; i++) {
            sum += mat[i][i];
            sum += mat[i][side - 1 - i];
        }
        if (side % 2 != 0)
            sum -= mat[side / 2][side / 2];
        return sum;
    }
}

  • // OJ: https://leetcode.com/problems/matrix-diagonal-sum/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int diagonalSum(vector<vector<int>>& A) {
        int N = A.size(), sum = 0;
        for (int i = 0; i < N; ++i) sum += i == N - i - 1 ? A[i][i] : (A[i][i] + A[N - i - 1][i]);
        return sum;
    }
};

  • class Solution:
    def diagonalSum(self, mat: List[List[int]]) -> int:
        n = len(mat)
        res = 0
        for i in range(n):
            res += mat[i][i] + (0 if n - i - 1 == i else mat[i][n - i - 1])
        return res

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