Given a positive integer N, the task is to print an N × N zig-zag matrix consisting of numbers from 1 to N2, such that the ZigZag traversal of the matrix yields the number in ascending order.
Examples:
Input: N = 3
Output:
1 2 4
3 5 7
6 8 9
Explanation:
Input: N = 4
Output:
1 2 4 7
3 5 8 11
6 9 12 14
10 13 15 16
Approach:
The required matrix can be broken down into two right-angled triangles.
- An upside-down right-angled triangle(considered as an upper triangle).
- A normal right-angled triangle(considered as a lower triangle).
The idea is to iterate two nested loops to fill the upper triangle with respective values. Then iterate two nested loop again to fill the lower triangle with respective values. After the above two operations print the desired matrix.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to print the pattern void printPattern( int n) { // N * N matrix to store the // values int arr[n][n]; arr[0][0] = 1; // Fill the values of // upper triangle for ( int i = 0; i < n; i++) { if (i > 0) { arr[i][0] = arr[i - 1][0] + i + 1; } for ( int j = 1; j < n - i; j++) { arr[i][j] = arr[i][j - 1] + i + j; } } // Fill the values of // lower triangle arr[1][n - 1] = arr[n - 1][0] + 1; int div = 0; for ( int i = 2; i < n; i++) { div = n - 2; for ( int j = n - i; j < n; j++) { if (j == n - i) { arr[i][j] = arr[i - 1][j + 1] + 1; } else { arr[i][j] = arr[i][j - 1] + div ; div --; } } } // Print the array for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) { cout << arr[i][j] << " " ; } cout << "\n" ; } } // Driver Code int main() { // Given size of matrix int N = 4; // Function Call printPattern(N); return 0; } |
Java
// Java program for // the above approach import java.util.*; class GFG{ // Function to print the pattern static void printPattern( int n) { // N * N matrix to store the // values int [][]arr = new int [n][n]; arr[ 0 ][ 0 ] = 1 ; // Fill the values of // upper triangle for ( int i = 0 ; i < n; i++) { if (i > 0 ) { arr[i][ 0 ] = arr[i - 1 ][ 0 ] + i + 1 ; } for ( int j = 1 ; j < n - i; j++) { arr[i][j] = arr[i][j - 1 ] + i + j; } } // Fill the values of // lower triangle arr[ 1 ][n - 1 ] = arr[n - 1 ][ 0 ] + 1 ; int div = 0 ; for ( int i = 2 ; i < n; i++) { div = n - 2 ; for ( int j = n - i; j < n; j++) { if (j == n - i) { arr[i][j] = arr[i - 1 ][j + 1 ] + 1 ; } else { arr[i][j] = arr[i][j - 1 ] + div; div--; } } } // Print the array for ( int i = 0 ; i < n; i++) { for ( int j = 0 ; j < n; j++) { System.out.print(arr[i][j] + " " ); } System.out.print( "\n" ); } } // Driver Code public static void main(String[] args) { // Given size of matrix int N = 4 ; // Function Call printPattern(N); } } // This code is contributed by Princi Singh |
Python3
# Python3 program for the above approach # Function to print the pattern def printPattern(n): # N * N matrix to store the values arr = [[ 0 for i in range (n)] for j in range (n)] # Fill the values of upper triangle arr[ 0 ][ 0 ] = 1 for i in range (n): if i > 0 : arr[i][ 0 ] = arr[i - 1 ][ 0 ] + i + 1 for j in range ( 1 , n - i): arr[i][j] = arr[i][j - 1 ] + i + j # Fill the values of lower triangle if n > 1 : arr[ 1 ][n - 1 ] = arr[n - 1 ][ 0 ] + 1 div = 0 for i in range ( 2 , n): div = n - 2 for j in range (n - i, n): if j = = n - i: arr[i][j] = arr[i - 1 ][j + 1 ] + 1 else : arr[i][j] = arr[i][j - 1 ] + div div - = 1 # Print the array for i in range (n): for j in range (n): print (arr[i][j], end = ' ' ) print ("") # Driver code # Given size of matrix N = 4 # Function Call printPattern(N) |
C#
// C# program for // the above approach using System; class GFG{ // Function to print the pattern static void printPattern( int n) { // N * N matrix to store the // values int [,]arr = new int [n, n]; arr[0,0] = 1; // Fill the values of // upper triangle for ( int i = 0; i < n; i++) { if (i > 0) { arr[i, 0] = arr[i - 1, 0] + i + 1; } for ( int j = 1; j < n - i; j++) { arr[i, j] = arr[i, j - 1] + i + j; } } // Fill the values of // lower triangle arr[1, n - 1] = arr[n - 1, 0] + 1; int div = 0; for ( int i = 2; i < n; i++) { div = n - 2; for ( int j = n - i; j < n; j++) { if (j == n - i) { arr[i, j] = arr[i - 1, j + 1] + 1; } else { arr[i, j] = arr[i, j - 1] + div; div--; } } } // Print the array for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) { Console.Write(arr[i, j] + " " ); } Console.Write( "\n" ); } } // Driver Code public static void Main(String[] args) { // Given size of matrix int N = 4; // Function Call printPattern(N); } } // This code is contributed by shikhasingrajput |
1 2 4 7 3 5 8 11 6 9 12 14 10 13 15 16
Time Complexity: O(N2)
Auxiliary Space: O(N2)
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