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++ 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 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 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# 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 |
<script> // Javascript program for // the above approach // Function to print the pattern function printPattern(n)
{ // N * N matrix to store the
// values
let arr = new Array(n);
// Loop to create 2D array using 1D array
for (let i = 0; i < arr.length; i++) {
arr[i] = new Array(2);
}
arr[0][0] = 1;
// Fill the values of
// upper triangle
for (let i = 0; i < n; i++)
{
if (i > 0)
{
arr[i][0] = arr[i - 1][0] +
i + 1;
}
for (let 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;
let div = 0;
for (let i = 2; i < n; i++)
{
div = n - 2;
for (let 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 (let i = 0; i < n; i++)
{
for (let j = 0; j < n; j++)
{
document.write(arr[i][j] + " " );
}
document.write( "<br/>" );
}
} // Driver code // Given size of matrix
let N = 4;
// Function Call
printPattern(N);
// This code is contributed by susmitakundugoaldanga.
</script> |
1 2 4 7 3 5 8 11 6 9 12 14 10 13 15 16
Time Complexity: O(N2)
Auxiliary Space: O(N2)