Added Solution1.cpp for Day-01 from Harjas

Author: harjasbb07-eng

Problems/Mathematics/Day-01/sol/Harjas/Solution1.cpp

@@ -0,0 +1,49 @@
+/*
+Problem Statement:
+Given three distinct integers a, b, and c placed on a circle of even size,
+numbered clockwise from 1. Person a is opposite to person b.
+Find the number opposite to c. If no such circle exists, return -1.
+
+Prefix Sums:
+Not applicable since the problem is purely mathematical and does not involve
+range queries or cumulative sums.
+
+Time Complexity: O(1)
+Space Complexity: O(1)
+
+Submission Link:
+https://codeforces.com/contest/1560/submission/355443029
+*/
+#include <bits/stdc++.h>
+using namespace std;
+
+int main(){
+
+int t;
+cin>>t;
+while(t--){
+int a,b,c,n;
+cin>>a>>b>>c;
+if(a>b) swap(a,b);    //naming the bigger number b and smaller one a
+
+int SizeOfCircle= 2*(b-a); // thinking: b and a form the diameter of circle 
+
+if(SizeOfCircle==2) cout<<-1<<endl; //edge case when a and b are adjacent
+else{
+if(c>SizeOfCircle|| b>SizeOfCircle) cout<<-1<<endl; // edge case when either c or the bigger number do not reside in the size
+else{
+int ans=0;
+if(c>b) ans=(c-(b-a));      // because a and b form diameter then c lies in upper or lower half and opposite of c also lies based on that
+else if(c<=b) ans=(c+(b-a));
+
+if(ans<=0) ans+=SizeOfCircle;
+else if(ans>SizeOfCircle) ans-=SizeOfCircle;   // fixing ans if it resides outside elements of circle
+
+cout<< ans<<endl;
+}
+
+
+}
+}
+return 0;
+}