Change/add only one character and print ‘*’ exactly 20 times
Last Updated :
02 Mar, 2023
In the below code, change/add only one character and print ‘*’ exactly 20 times.
int main()
{
int i, n = 20;
for (i = 0; i < n; i--)
printf("*");
getchar();
return 0;
}
Solutions:
1. Replace i by n in for loop’s third expression
C++
#include <iostream>
using namespace std;
int main()
{
int i, n = 20;
for (i = 0; i < n; n--)
cout << "*";
getchar();
return 0;
}
|
C
#include <stdio.h>
int main()
{
int i, n = 20;
for (i = 0; i < n; n--)
printf("*");
getchar();
return 0;
}
|
Java
class GfG {
public static void main(String[] args)
{
int i, n = 20;
for (i = 0; i < n; n--)
System.out.print("*");
}
}
|
Python3
if __name__ == '__main__':
n = 20
for i in range(0, n):
print("*",end='')
n -= 1
|
C#
using System;
class GfG
{
public static void Main()
{
int i, n = 20;
for (i = 0; i < n; n--)
Console.Write("*");
}
}
|
Javascript
<script>
var i, n = 20;
for(i = 0; i < n; n--)
document.write("*");
</script>
|
Output
********************
2. Put ‘-‘ before i in for loop’s second expression
C++
#include<bits/stdc++.h>
using namespace std;
int main()
{
int i, n = 20;
for (i = 0; -i < n; i--)
cout<<"*";
return 0;
}
|
C
#include <stdio.h>
int main()
{
int i, n = 20;
for (i = 0; -i < n; i--)
printf("*");
getchar();
return 0;
}
|
Java
import java.util.*;
public class GFG
{
public static void main(String[] args)
{
int i, n = 20;
for (i = 0; -i < n; i--)
System.out.print("*");
}
}
|
Python3
if __name__ == '__main__':
n = 20
for i in range(0,n):
print("*", end="")
|
C#
using System;
class GfG
{
public static void Main()
{
int i, n = 20;
for (i = 0; -i < n; i--)
Console.Write("*");
}
}
|
Javascript
<script>
let i, n = 20;
for (i = 0; -i < n; i--)
document.write("*");
</script>
|
Output
********************
3. Replace < by + in for loop’s second expression
C++
#include <iostream>
using namespace std;
int main()
{
int i, n = 20;
for(i = 0; i + n; i--)
cout << "*";
return 0;
}
|
C
#include <stdio.h>
int main()
{
int i, n = 20;
for (i = 0; i + n; i--)
printf("*");
getchar();
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void main (String[] args) {
int i, n = 20;
for(i = 0; i + n > 0; i--)
System.out.print("*");
}
}
|
Python3
i,n = 0,20
while(i+n):
print("*",end="")
i -= 1
|
C#
using System;
class GFG
{
static void Main(string[] args)
{
int i, n = 20;
for (i = 0; i + n > 0; i--)
Console.Write("*");
}
}
|
Javascript
<script>
let i, n = 20;
for(i = 0; i + n; i--)
document.write("*");
</script>
|
Output
********************
Let’s extend the problem little.
Change/add only one character and print ‘*’ exactly 21 times.
Solution: Put negation operator before i in for loop’s second expression.
Explanation: Negation operator converts the number into its one’s complement.
No. One's complement
0 (00000..00) -1 (1111..11)
-1 (11..1111) 0 (00..0000)
-2 (11..1110) 1 (00..0001)
-3 (11..1101) 2 (00..0010)
...............................................
-20 (11..01100) 19 (00..10011)
C++
#include <iostream>
using namespace std;
int main()
{
int i, n = 20;
for (i = 0; ~i < n; i--)
printf("*");
getchar();
return 0;
}
|
C
#include <stdio.h>
int main()
{
int i, n = 20;
for (i = 0; ~i < n; i--)
printf("*");
getchar();
return 0;
}
|
Java
import java.util.*;
class GFG{
public static void main(String[] args)
{
int i, n = 20;
for(i = 0; ~i < n; i--)
System.out.print( "*" );
}
}
|
Python3
n = 20
i = 0
while(~i<n):
print("*",end="")
i -= 1
|
C#
using System;
class GFG{
public static void Main(String[] args)
{
int i, n = 20;
for(i = 0; ~i < n; i--)
Console.Write( "*" );
}
}
|
Javascript
<script>
{
var i, n = 20;
for(i = 0; ~i < n; i--)
document.write("*");
}
</script>
|
Output
*********************
Time Complexity: O(1)
Since only a single loop is used to print the n stars, the time complexity of this approach is O(1).
Space Complexity: O(1)
No extra space is used in this approach and hence the space complexity is O(1).
Please comment if you find more solutions to the above problems.
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