Print lower triangle with alternate ‘*’ and ‘#’
Last Updated :
14 Mar, 2023
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).
- After the outer loop has been completed, the pattern has been printed.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n = 6;
for (int row = 1; row <= n; row++) {
for (int col = 1; col <= row; col++) {
if (col % 2 == 0) {
cout << "#";
}
else {
cout << "*";
}
}
cout << endl;
}
return 0;
}
|
Python3
n = 6
for row in range(1, n+1):
for col in range(1, row+1):
if col % 2 == 0:
print("#", end="")
else:
print("*", end="")
print()
|
C#
using System;
public class Gfg
{
public static void Main(string[] args)
{
int n = 6;
for (int row = 1; row <= n; row++) {
for (int col = 1; col <= row; col++) {
if (col % 2 == 0) {
Console.Write("#");
}
else {
Console.Write("*");
}
}
Console.Write("\n");
}
}
}
|
Javascript
let n = 6;
for (let row = 1; row <= n; row++) {
for (let col = 1; col <= row; col++) {
if (col % 2 == 0) {
console.log("#");
}
else {
console.log("*");
}
}
console.log("<br>");
}
|
Java
import java.util.*;
class GFG{
public static void main(String args[])
{
int n = 6;
for (int row = 1; row <= n; row++) {
for (int col = 1; col <= row; col++) {
if (col % 2 == 0) {
System.out.print("#");
}
else {
System.out.print("*");
}
}
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.
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