README_EN.md

1572. Matrix Diagonal Sum

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Description

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

Solutions

Python3

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

Java

class Solution {
    public int diagonalSum(int[][] mat) {
        int n = mat.length;
        int res = 0;
        for (int i = 0; i < n; ++i) {
            res += mat[i][i] + (n - i - 1 == i ? 0 : mat[i][n - i - 1]);
        }
        return res;
    }
}

TypeScript

function diagonalSum(mat: number[][]): number {
    let n = mat.length;
    let ans = 0;
    for (let i = 0; i < n; i++) {
        ans += mat[i][i];
        let j = n - 1 - i;
        if (i != j) {
            ans += mat[i][j];
        }
    }
    return ans;
}

C++

class Solution {
public:
    int diagonalSum(vector<vector<int>>& mat) {
        int n = mat.size();
        int res = 0;
        for (int i = 0; i < n; ++i) {
            res += mat[i][i] + (n - i - 1 == i ? 0 : mat[i][n - i - 1]);
        }
        return res;
    }
};

Go

func diagonalSum(mat [][]int) int {
	n, res := len(mat), 0
	for i := 0; i < n; i++ {
		res += mat[i][i]
		if n-i-1 != i {
			res += mat[i][n-i-1]
		}
	}
	return res
}

Rust

impl Solution {
    pub fn diagonal_sum(mat: Vec<Vec<i32>>) -> i32 {
        let n = mat.len();
        let mut res = 0;
        for i in 0..n {
            res += mat[i][i] + mat[n - i - 1][i];
        }
        if n & 1 == 1 {
            return res - mat[n / 2][n / 2];
        }
        res
    }
}

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