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
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