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
Implementation: Following is the C++ implementation for the above pattern.
C++
// 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;} |
Java
// JAVA program to print // matrix downwardclass GFG{static void printMatrixDiagonallyDown(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) { System.out.print(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) { System.out.print(matrix[row][col] + " "); row++; col--; } }}// Driver codepublic static void main(String[] args){ int[][] matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; int n = 3; printMatrixDiagonallyDown(matrix, n);}}// This code is contributed by Rajput-Ji |
Python 3
# Python 3 program to print matrix downwarddef 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 -= 1if __name__ == '__main__': matrix = [[1, 2, 3],[4, 5, 6],[7, 8, 9]] n = 3 printMatrixDiagonallyDown(matrix, n)# This code is contributed by Surendra_Gangwar |
C#
// C# program to print // matrix downwardusing System;class GFG{static void printMatrixDiagonallyDown(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) { Console.Write(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) { Console.Write(matrix[row,col] + " "); row++; col--; } }}// Driver codepublic static void Main(String[] args){ int[,] matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; int n = 3; printMatrixDiagonallyDown(matrix, n);}}// This code is contributed by Amit Katiyar |
Javascript
<script>// JavaScript program to print// matrix downwardfunction printMatrixDiagonallyDown(matrix,n){// printing elements above and on// second diagonalfor (let k = 0; k < n; k++){ // traversing downwards // starting from first row let row = 0, col = k; while (col >= 0) { document.write(matrix[row][col] + " "); row++; col--; }}// printing elements below// second diagonalfor (let j = 1; j < n; j++){ // traversing downwards starting // from last column let col = n - 1, row = j; while (row < n) { document.write(matrix[row][col] + " "); row++; col--; }}}// Driver codelet matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]];let n = 3;printMatrixDiagonallyDown(matrix, n);// This code is contributed by sravan kumar</script> |
Output
1 2 4 3 5 7 6 8 9
Complexity Analysis:
- Time Complexity: O(n2)
- Auxiliary Space: 1
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!




