Print the matrix diagonally downwards

Given a matrix of size n*n, print the matrix in the following pattern.


Output: 1 2 5 3 6 9 4 7 10 13 8 11 14 12 15 16

Examples:

Input :matrix[2][2]= { {1, 2},
                       {3, 4} }
Output : 1 2 3 4

Input :matrix[3][3]= { {1, 2, 3},
                       {4, 5, 6},
                       {7, 8, 9} }
Output : 1 2 4 3 5 7 6 8 9

Following is the C++ implementation for the above pattern.

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// CPP program to print matrix downward
#include <bits/stdc++.h>
using namespace std;
  
void printMatrixDiagonallyDown(vector<vector<int> > matrix, 
                                                    int n)
{
    // printing elements above and on 
    // second diagonal
    for (int k = 0; k < n; k++) {
  
        // traversing downwards starting
        // from first row
        int row = 0, col = k;
        while (col >= 0) {
            cout << matrix[row][col] << " ";
            row++, col--;
        }
    }
  
    // printing elements below second
    // diagonal
    for (int j = 1; j < n; j++) {
  
        // traversing downwards starting 
        // from last column
        int col = n - 1, row = j;
        while (row < n) {
            cout << matrix[row][col] << " ";
            row++, col--;
        }
    }
}
  
int main()
{
    vector<vector<int> > matrix{ { 1, 2, 3 },
                                 { 4, 5, 6 },
                                 { 7, 8, 9 } };
    int n = 3;
    printMatrixDiagonallyDown(matrix, n);
    return 0;
}
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# Python 3 program to print matrix downward
  
def printMatrixDiagonallyDown(matrix,n):
    # printing elements above and on 
    # second diagonal
    for k in range(n):
        # traversing downwards starting
        # from first row
        row = 0
        col = k
        while (col >= 0):
            print(matrix[row][col],end = " ")
            row += 1
            col -= 1
  
    # printing elements below second
    # diagonal
    for j in range(1,n):
        # traversing downwards starting 
        # from last column
        col = n - 1
        row = j
        while (row < n):
            print(matrix[row][col],end = " ")
            row += 1
            col -= 1
  
if __name__ == '__main__':
    matrix = [[1, 2, 3],[4, 5, 6],[7, 8, 9]]
    n = 3
    printMatrixDiagonallyDown(matrix, n)
  
# This code is contributed by Surendra_Gangwar
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Output:

1 2 4 3 5 7 6 8 9

Time Complexity: O(n*n)

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Improved By : pqrss, SURENDRA_GANGWAR

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