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;
}
// CPP program to print double quotes
#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
// C program to check if two numbers are equal
// without using arithmetic operators or
// comparison operators
#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++ program to check if two numbers are equal
// without using arithmetic operators or
// comparison operators
#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;
}
// This code is contributed by shivani
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++ program to print all natural numbers upto
// N without using semi-colon
#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++ program to check if two numbers are equal
#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++ program to find maximum and minimum of
// two numbers without using loop and any
// condition.
#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
// C program to print maximum value of
// unsigned int.
#include<stdio.h>
int main()
{
unsigned int max;
max = 0;
max = ~max;
printf("Max value : %u ", max);
return 0;
}
// C++ program to print maximum value of
// unsigned int.
#include <iostream>
int main()
{
unsigned int max;
max = 0;
max = ~max;
std::cout << "Max value : " << max;
return 0;
}
// This code is contributed by sarajadhav12052009
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;
}
// CPP program to print sum of two integers
// without +
#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;
}
// CPP program to verifies the condition inside if block
// It just verifies the condition inside if block,
// i.e., cout << "geeks" which returns a non-zero value,
// !(non-zero value) is false, hence it executes else
// Hence technically it only executes else block
#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++
// CPP program to divide a number by 4
// without using '/'
#include<iostream>
using namespace std;
int main()
{
int n = 4;
n = n >> 2;
cout << n;
return 0;
}
#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
// C program to find if machine is little
// endian or big endian.
#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++ program to find if machine is little
// endian or big endian.
#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;
}
// This code is contributed by sarajadhav12052009
Time Complexity: O(1)
Auxiliary Space: O(1)
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