Given three integers A, B and C. In an infinite sequence, A is the first number, C is the common difference (Si – Si – 1 = C). The task is to check if the number B will appear in the sequence or not.
Examples:
Input: A = 1, B = 7, C = 3
Output: Yes
The sequence will be 1, 4, 7, 10, …
Input: A = 1, B = -4, C = 5
Output: No
Approach: There are two cases:
- When C = 0, print Yes if A = B else No as the sequence will consist only the number A
- When C > 0, for any non-negative integer k the equation B = A + k * C must be satisfied i.e. (B – A) / C must be a non-negative integer.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function that returns true if // the sequence will contain B bool doesContainB( int a, int b, int c)
{ if (a == b)
return true ;
if ((b - a) * c > 0 && (b - a) % c == 0)
return true ;
return false ;
} // Driver code int main()
{ int a = 1, b = 7, c = 3;
if (doesContainB(a, b, c))
cout << "Yes" ;
else
cout << "No" ;
return 0;
} |
Java
// Java implementation of the approach class GFG
{ // Function that returns true if
// the sequence will contain B
static boolean doesContainB( int a, int b, int c)
{
if (a == b)
{
return true ;
}
if ((b - a) * c > 0 && (b - a) % c == 0 )
{
return true ;
}
return false ;
}
// Driver code
public static void main(String[] args)
{
int a = 1 , b = 7 , c = 3 ;
if (doesContainB(a, b, c))
{
System.out.println( "Yes" );
}
else
{
System.out.println( "No" );
}
}
} // This code contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach # Function that returns true if # the sequence will contain B def doesContainB(a, b, c):
if (a = = b):
return True
if ((b - a) * c > 0 and (b - a) % c = = 0 ):
return True
return False
# Driver code if __name__ = = '__main__' :
a, b, c = 1 , 7 , 3
if (doesContainB(a, b, c)):
print ( "Yes" )
else :
print ( "No" )
# This code is contributed by 29AjayKumar |
C#
// C# implementation of the approach using System;
class GFG
{ // Function that returns true if
// the sequence will contain B
static bool doesContainB( int a, int b, int c)
{
if (a == b)
{
return true ;
}
if ((b - a) * c > 0 && (b - a) % c == 0)
{
return true ;
}
return false ;
}
// Driver code
public static void Main()
{
int a = 1, b = 7, c = 3;
if (doesContainB(a, b, c))
{
Console.WriteLine( "Yes" );
}
else
{
Console.WriteLine( "No" );
}
}
} /* This code contributed by PrinciRaj1992 */ |
PHP
<?php // PHP implementation of the approach // Function that returns true if // the sequence will contain B function doesContainB( $a , $b , $c )
{ if ( $a == $b )
return true;
if (( $b - $a ) * $c > 0 &&
( $b - $a ) % $c == 0)
return true;
return false;
} // Driver code $a = 1; $b = 7; $c = 3;
if (doesContainB( $a , $b , $c ))
echo "Yes" ;
else echo "No" ;
// This code is contributed // by Akanksha Rai ?> |
Javascript
<script> // javascript program for the above approach // Function that returns true if
// the sequence will contain B
function doesContainB(a, b, c)
{
if (a == b)
{
return true ;
}
if ((b - a) * c > 0 && (b - a) % c == 0)
{
return true ;
}
return false ;
}
// Driver Code let a = 1, b = 7, c = 3;
if (doesContainB(a, b, c))
{
document.write( "Yes" );
}
else
{
document.write( "No" );
}
</script> |
Output:
Yes
Time Complexity: O(1), since there is only basic arithmetic which takes constant time.
Auxiliary Space: O(1), since no extra space has been taken.