Find K such that A – K = B – K
Given two integers A and B where (A ≠ B). The task is to find K such that A – K = B – K. If no such K exists then print 1.
Examples:
Input: A = 2, B = 16
Output: 9
2 – 9 = 16 – 9 = 7
Input: A = 5, B = 2
Output: 1
Approach: It is given that A ≠ B. So let A < B then there are three cases:
 K < A: This gives A – K = B – K which gives A = B which is false.
 K > B: This gives K – A = K – B which is also false.
 A ≤ K ≤ B: This gives K – A = B – K which gives 2 * K = A + B
If A + B is odd, there is thus no solution. If A + B is even then the answer is (A + B) / 2.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to find k such that // a  k = b  k int find_k( int a, int b) { // If (a + b) is even if ((a + b) % 2 == 0) return ((a + b) / 2); return 1; } // Driver code int main() { int a = 2, b = 16; cout << find_k(a, b); return 0; } 
Java
// Java implementation of the approach class GFG { // Function to find k such that // a  k = b  k static int find_k( int a, int b) { // If (a + b) is even if ((a + b) % 2 == 0 ) return ((a + b) / 2 ); return  1 ; } // Driver code public static void main(String[] args) { int a = 2 , b = 16 ; System.out.println(find_k(a, b)); } } // This code is contributed by Code_Mech 
Python3
# Python3 implementation of the approach # Function to find k such that # a  k = b  k def find_k(a, b) : # If (a + b) is even if ((a + b) % 2 = = 0 ) : return ((a + b) / / 2 ); return  1 ; # Driver code if __name__ = = "__main__" : a = 2 ; b = 16 ; print (find_k(a, b)); # This code is contributed by AnkitRai01

9
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