Given a number N which denotes the number of rows, the task is to follow the below pattern to print the first N rows of it.
Pattern:
*
*#
*#*
*#*#
*#*#*
Examples:
Input: N = 2
Output:
*
*#Input: N = 6
Output:
*
*#
*#*
*#*#
*#*#*
*#*#*#
Approach: Follow the below steps to implement the above pattern:
- Initialize two variables row and col to 1. These will be used to keep track of the current row and column that we are on in the pattern.
- Use a loop to iterate the row from 1 to N. This will be the outer loop and will represent each row of the pattern.
- Inside the outer loop, use another loop to iterate col from 1 to row. This will be the inner loop and will represent each column in the current row.
- Inside the inner loop, check if col is even or odd.
- If it is even, print a “#” character. If it is odd, print a “*” character.
- After the inner loop has been completed, move to the next line (this will start a new row in the pattern).
- Inside the outer loop, use another loop to iterate col from 1 to row. This will be the inner loop and will represent each column in the current row.
- After the outer loop has been completed, the pattern has been printed.
Below is the implementation of the above approach:
C++
// C++ code for the above approach: #include <bits/stdc++.h> using namespace std;
int main()
{ int n = 6;
// Loop through each row of
// the pattern
for ( int row = 1; row <= n; row++) {
// Loop through each column of
// the pattern
for ( int col = 1; col <= row; col++) {
// If the column number is even,
// print a "#" character
if (col % 2 == 0) {
cout << "#" ;
}
// If the column number is odd,
// print a "*" character
else {
cout << "*" ;
}
}
// Move to the next line after
// printing each row
cout << endl;
}
return 0;
} |
Python3
# Python code for the above approach: n = 6
# Loop through each row of # the pattern for row in range ( 1 , n + 1 ):
# Loop through each column of
# the pattern
for col in range ( 1 , row + 1 ):
# If the column number is even,
# print a "#" character
if col % 2 = = 0 :
print ( "#" , end = "")
# If the column number is odd,
# print a "*" character
else :
print ( "*" , end = "")
# Move to the next line after
# printing each row
print ()
|
C#
// C# code for the above approach: using System;
public class Gfg
{ public static void Main( string [] args)
{
int n = 6;
// Loop through each row of
// the pattern
for ( int row = 1; row <= n; row++) {
// Loop through each column of
// the pattern
for ( int col = 1; col <= row; col++) {
// If the column number is even,
// print a "#" character
if (col % 2 == 0) {
Console.Write( "#" );
}
// If the column number is odd,
// print a "*" character
else {
Console.Write( "*" );
}
}
// Move to the next line after
// printing each row
Console.Write( "\n" );
}
}
} // This code is contributed by ritaagarwal. |
Javascript
// Javascript code for the above approach: let n = 6; // Loop through each row of // the pattern for (let row = 1; row <= n; row++) {
// Loop through each column of
// the pattern
for (let col = 1; col <= row; col++) {
// If the column number is even,
// print a "#" character
if (col % 2 == 0) {
console.log( "#" );
}
// If the column number is odd,
// print a "*" character
else {
console.log( "*" );
}
}
// Move to the next line after
// printing each row
console.log( "<br>" );
} // This code is contributed by poojaagarwal2. |
Java
// Java program for the above approach import java.util.*;
class GFG{
public static void main(String args[])
{
int n = 6 ;
// Loop through each row of
// the pattern
for ( int row = 1 ; row <= n; row++) {
// Loop through each column of
// the pattern
for ( int col = 1 ; col <= row; col++) {
// If the column number is even,
// print a "#" character
if (col % 2 == 0 ) {
System.out.print( "#" );
}
// If the column number is odd,
// print a "*" character
else {
System.out.print( "*" );
}
}
// Move to the next line after
// printing each row
System.out.print( "\n" );
}
}
} |
Output
* *# *#* *#*# *#*#* *#*#*#
Time Complexity: O(N2) // since two nested loops are used the time taken by the algorithm to complete all operations is quadratic.
Auxiliary Space: O(1) //since there is a basic arithmetic operation that takes constant time.
Article Tags :
Recommended Articles