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

Last Updated : 09 Oct, 2020

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

Output

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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