anti-diagonals.md

Anti Diagonals

Problem Description

Give a N * N square matrix A, return an array of its anti-diagonals. Look at the example for more details.

Problem Constraints

1<= N <= 1000
1<= A[i][j] <= 1e9

Input Format

Only argument is a 2D array A of size N * N.

Output Format

Return a 2D integer array of size (2 * N-1) * N, representing the anti-diagonals of input array A.
The vacant spaces in the grid should be assigned to 0.

Example Input

Input 1:
1 2 3
4 5 6
7 8 9

Input 2:
1 2
3 4

Example Output

Output 1:
1 0 0
2 4 0
3 5 7
6 8 0
9 0 0

Output 2:
1 0
2 3
4 0

Example Explanation

Explanation 1:
The first anti diagonal of the matrix is [1 ], the rest spaces shoud be filled with 0 making the row as [1, 0, 0].
The second anti diagonal of the matrix is [2, 4 ], the rest spaces shoud be filled with 0 making the row as [2, 4, 0].
The third anti diagonal of the matrix is [3, 5, 7 ], the rest spaces shoud be filled with 0 making the row as [3, 5, 7].
The fourth anti diagonal of the matrix is [6, 8 ], the rest spaces shoud be filled with 0 making the row as [6, 8, 0].
The fifth anti diagonal of the matrix is [9 ], the rest spaces shoud be filled with 0 making the row as [9, 0, 0].

Explanation 2:
The first anti diagonal of the matrix is [1 ], the rest spaces shoud be filled with 0 making the row as [1, 0, 0].
The second anti diagonal of the matrix is [2, 4 ], the rest spaces shoud be filled with 0 making the row as [2, 4, 0].
The third anti diagonal of the matrix is [3, 0, 0 ], the rest spaces shoud be filled with 0 making the row as [3, 0, 0].

Output

function antiDiagonals(A){
        const N = A.length
        const M = A[0].length

        let result = []

        if(N == 0 || M == 0){
            return
        }

        for(let c = 0 ; c < M ; c++){
            let i = 0;
            let j = c;
            let interArr = []
            while(i < N && j >= 0){
                interArr.push(A[i][j])
                i++;
                j--;
            }
            while(interArr.length < N){
                interArr.push(0)
            }

            result.push(interArr)
        }

        for(let r = 1 ; r < N ; r++){
            let i = r;
            let j = M - 1;
            let interArr = []
            while(i < N && j >= 0){
                interArr.push(A[i][j])
                i++;
                j--;
            }
            while(interArr.length < N){
                interArr.push(0)
            }

            result.push(interArr)
        }

        return result
}