Given the value of n i.e, the number of rows, print the following pattern.
Examples :
Input : n = 4 Output : 1 5 2 8 6 3 10 9 7 4 Input : n = 6 Output : 1 7 2 12 8 3 16 13 9 4 19 17 14 10 5 21 20 18 15 11 6
Approach: The approach is to start printing the pattern from the end of each row. After the completion of the last column of each row, start from the second last column of the second row and so on. Below given is the implementation of this approach :
C++
// C++ program to print the pattern #include <bits/stdc++.h> using namespace std; // Function to print pattern // for given value of n void pattern(int n) { int p, k = 1; // Outer "for" loop for number of rows for (int i = 1; i <= n; i++) { // set the value of "p" as "k". p = k; // Inner "for" loop for number of columns for (int j = 1; j <= i; j++) { // print the values cout << p << " "; // change in value of "p" for // every elements after the first // element of each row . p = p - (n + j - i); } cout << endl; // value of "k" for first // element of every row. k = k + 1 + n - i; } } // Driver Code int main() { int n = 5; // Function calling pattern(n); return 0; } |
Java
// Java program to print the given pattern import java.util.*; class GfG { // Function to print pattern for // given value of n static void pattern(int n) { int p, k = 1; // Outer "for" loop for number of rows for (int i = 1; i <= n; i++) { // set the value of "p" as "k". p = k; // Inner "for" loop for number of columns for (int j = 1; j <= i; j++) { // print the values System.out.print(p); System.out.print(" "); // change in value of "p" for // every elements after the // first element of each row . p = p - (n + j - i); } // Print the next line System.out.println(); // value of "k" for first // element of every row. k = k + 1 + n - i; } } // Driver Code public static void main(String[] args) { int n = 5; // Function calling pattern(n); } } |
Python 3
# Python 3 program to print the pattern # Function to print pattern # for given value of n def pattern(n): k = 1 # Outer "for" loop for number of rows for i in range(1 , n+1): # set the value of "p" as "k". p = k # Inner "for" loop for number of columns for j in range(1 , i+1): # print the values print(p , end=" ") # change in value of "p" for # every elements after the first # element of each row . p = p - (n + j - i) print("") # value of "k" for first # element of every row. k = k + 1 + n - i # Driver Code n = 5 # Function calling pattern(n) # This code is contributed by Smitha |
C#
// C# program to print the given pattern using System; class GfG { // Function to print pattern for // given value of n static void pattern(int n) { int p, k = 1; // Outer "for" loop for number // of rows for (int i = 1; i <= n; i++) { // set the value of "p" as "k". p = k; // Inner "for" loop for number // of columns for (int j = 1; j <= i; j++) { // print the values Console.Write(p); Console.Write(" "); // change in value of "p" for // every elements after the // first element of each row . p = p - (n + j - i); } // Print the next line Console.WriteLine(); // value of "k" for first // element of every row. k = k + 1 + n - i; } } // Driver Code public static void Main() { int n = 5; // Function calling pattern(n); } } // This code is contributed by anuj_67. |
PHP
<?php // PHP program to print the pattern // Function to print pattern for // given value of n function pattern($n) { $p; $k = 1; // Outer "for" loop for number of rows for($i = 1; $i <= $n; $i++) { // set the value of "p" as "k". $p = $k; // Inner "for" loop for number of columns for($j = 1; $j <= $i; $j++) { // print the values echo $p; echo " "; // change in value of "p" for // every elements after the // first element of each row . $p = $p - ($n + $j - $i); } echo "\n"; // value of "k" for first // element of every row. $k = $k + 1 + $n - $i; } } // Driver Code $n = 5; // Function calling pattern($n); ?> |
Javascript
<script> // JavaScript program to print the pattern // Function to print pattern // for given value of n function pattern(n) { var p, k = 1; // Outer "for" loop for number of rows for (var i = 1; i <= n; i++) { // set the value of "p" as "k". p = k; // Inner "for" loop for number of columns for (var j = 1; j <= i; j++) { // print the values document.write(p + " "); // change in value of "p" for // every elements after the first // element of each row . p = p - (n + j - i); } document.write("<br>"); // value of "k" for first // element of every row. k = k + 1 + n - i; } } // Driver Code var n = 5; // Function calling pattern(n); </script> |
Output
1 6 2 10 7 3 13 11 8 4 15 14 12 9 5
Time Complexity: O(n2)
Auxiliary Space: O(1)
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