1572. Matrix Diagonal Sum

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

Solution 1

Python Code
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class Solution:
    def diagonalSum(self, mat: List[List[int]]) -> int:
        ans = 0
        n = len(mat)
        for i, row in enumerate(mat):
            j = n - i - 1
            ans += row[i] + (0 if j == i else row[j])
        return ans

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

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

Go Code
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func diagonalSum(mat [][]int) (ans int) {
	n := len(mat)
	for i, row := range mat {
		ans += row[i]
		if j := n - i - 1; j != i {
			ans += row[j]
		}
	}
	return
}

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

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

C Code
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int diagonalSum(int** mat, int matSize, int* matColSize) {
    int ans = 0;
    for (int i = 0; i < matSize; i++) {
        ans += mat[i][i] + mat[i][matSize - 1 - i];
    }
    if (matSize & 1) {
        ans -= mat[matSize >> 1][matSize >> 1];
    }
    return ans;
}

Solution 2

TypeScript Code
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function diagonalSum(mat: number[][]): number {
    const n = mat.length;
    let ans = 0;
    for (let i = 0; i < n; i++) {
        ans += mat[i][i] + mat[i][n - 1 - i];
    }
    if (n & 1) {
        ans -= mat[n >> 1][n >> 1];
    }
    return ans;
}