Arrange numbers 1 to N^2 in a Zig-Zag Matrix in ascending order

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 angle triangles.

An upside down right angled triangle(considered as upper triangle).
A normal right angled triangle(considered as lower triangle).

The idea is to iterate two nested loop 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; 
} 

Output:

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

Time Complexity: O(N²)
Auxillary Space: O(N²)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

C++

Recommended Posts: