We may come across various tricky programs in our day-to-day life. Maybe in technical interviews, coding tests, or C/C++ classrooms.
Here is a list of such programs:-
- Print text within double quotes (” “).
This may seem easy, but beginners may get puzzled while printing text within double quotes.
C
#include <stdio.h>
int main()
{
printf("\"geeksforgeeks\"");
return 0;
}
|
C++
#include<iostream>
int main()
{
std::cout << "\"geeksforgeeks\"";
return 0;
}
|
Time Complexity: O(1)
Auxiliary Space: O(1)
- To check if two numbers are equal without using arithmetic operators or comparison operators.
The simplest solution for this is using the Bitwise XOR operator (^). We know that for two equal numbers XOR operator returns 0. We will use the XOR operator to solve this problem.
C
#include<stdio.h>
int main()
{
int x = 10;
int y = 10;
if ( !(x ^ y) )
printf(" x is equal to y ");
else
printf(" x is not equal to y ");
return 0;
}
|
C++
#include <iostream>
using namespace std;
int main()
{
int x = 10;
int y = 10;
if (!(x ^ y))
cout << " x is equal to y ";
else
cout << " x is not equal to y ";
return 0;
}
|
Time Complexity: O(1)
Auxiliary Space: O(1)
- Print all natural numbers up to N without using a semi-colon.
We use the idea of recursively calling the main function.
C
#include<stdio.h>
int N = 10;
int main()
{
static int x = 1;
if (printf("%d ", x) && x++ < N && main())
{ }
return 0;
}
|
C++
#include<iostream>
using namespace std;
int N = 10;
int main()
{
static int x = 1;
if (cout << x << " " && x++ < N && main())
{ }
return 0;
}
|
Output1 2 3 4 5 6 7 8 9 10
Time Complexity: O(1)
Auxiliary Space: O(1)
- To Swap the values of two variables without using any extra variable.
C
#include<stdio.h>
int main()
{
int x = 10;
int y = 70;
x = x + y;
y = x - y;
x = x - y;
printf("X : %d\n", x);
printf("Y : %d\n", y);
return 0;
}
|
C++
#include<bits/stdc++.h>
using namespace std;
int main()
{
int x = 10;
int y = 70;
x = x + y;
y = x - y;
x = x - y;
cout << "X : " << x << "\n";
cout << "Y : " << y << "\n";
return 0;
}
|
Time Complexity: O(1)
Auxiliary Space: O(1)
- Program to find the Maximum and minimum of two numbers without using any loop or condition.
The simplest trick is-
C
#include<stdio.h>
#include<stdlib.h>
int main()
{
int a = 15, b = 20;
printf("max = %d\n", ((a + b) + abs(a - b)) / 2);
printf("min = %d", ((a + b) - abs(a - b)) / 2);
return 0;
}
|
C++
#include<bits/stdc++.h>
int main ()
{
int a = 15, b = 20;
printf("max = %d\n", ((a + b) + abs(a - b)) / 2);
printf("min = %d", ((a + b) - abs(a - b)) / 2);
return 0;
}
|
Time Complexity: O(1)
Auxiliary Space: O(1)
- Print the maximum value of an unsigned int using One’s Complement (~) Operator in C.
Here is a trick to find the maximum value of an unsigned int using one’s complement operator:
C
#include<stdio.h>
int main()
{
unsigned int max;
max = 0;
max = ~max;
printf("Max value : %u ", max);
return 0;
}
|
C++
#include <iostream>
int main()
{
unsigned int max;
max = 0;
max = ~max;
std::cout << "Max value : " << max;
return 0;
}
|
OutputMax value : 4294967295
Time Complexity: O(1)
Auxiliary Space: O(1)
- To find the sum of two integers without using ‘+’ operator.
This is a very easy mathematics trick.
We know that a + b = – (-a-b). So this will work as a trick for us.
C
#include <stdio.h>
int main()
{
int a = 5;
int b = 5;
int sum = -( -a-b );
printf("%d",sum);
return 0;
}
|
C++
#include<iostream>
using namespace std;
int main()
{
int a = 5;
int b = 5;
int sum = -( -a-b );
cout << sum;
return 0;
}
|
Time Complexity: O(1)
Auxiliary Space: O(1)
- Program to verify the condition inside if block.
C
#include <stdio.h>
int main()
{
if (!(printf("geeks")))
printf(" geeks ");
else
printf("forgeeks ");
return 0;
}
|
C++
#include<iostream>
using namespace std;
int main()
{
if (!(cout << "geeks"))
cout <<" geeks ";
else
cout << "forgeeks ";
return 0;
}
|
Time Complexity: O(1)
Auxiliary Space: O(1)
- Program to divide an integer by 4 without using ‘/’ operator.
One of the most efficient ways to divide an integer by 4 is to use right shift operator (“>>”).
C++
#include<iostream>
using namespace std;
int main()
{
int n = 4;
n = n >> 2;
cout << n;
return 0;
}
|
C
#include <stdio.h>
int main()
{
int n = 4;
n = n >> 2;
printf(" %d ",n);
return 0;
}
|
Time Complexity: O(1)
Auxiliary Space: O(1)
- Program to check endianness of the computer.
C
#include <stdio.h>
int main()
{
unsigned int n = 1;
char *c = (char*)&n;
if (*c)
printf("LITTLE ENDIAN");
else
printf("BIG ENDIAN");
return 0;
}
|
C++
#include <iostream>
int main()
{
unsigned int n = 1;
char *c = (char*)&n;
if (*c)
std::cout << "LITTLE ENDIAN";
else
std::cout << "BIG ENDIAN";
return 0;
}
|
Time Complexity: O(1)
Auxiliary Space: O(1)
This article is contributed by Smitha Dinesh Semwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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