Day-01 q001 solved

Author: mannatjain11465-netizen

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

@@ -0,0 +1,38 @@
+/*
+Problem Statement: 
+There are n people standing in a circle (n is even). Each person looks at the person standing exactly opposite to them. Given three distinct integers a, b, and c: Person a is looking at person b. Find the person that person c is looking at. If no valid circle can be formed, print -1.
+
+Approach:
+If person a looks at b, then the total number of people in the circle is:
+n = 2 * |a - b| For person c, if c + n/2 ≤ n, answer = c + n/2, else, answer = c - n/2 If n is zero or any of a, b, or c exceeds n, the configuration is invalid.
+
+Time Complexity:
+O(1)
+
+Space Complexity:
+O(1)*/
+
+#include <bits/stdc++.h>
+using namespace std;
+int main(){
+    int t;
+    cin>>t;
+    while(t--){
+        long a,b,c;
+        cin>>a>>b>>c;
+        long n=2*abs(a-b);
+        long ans=0;
+        if(n==0||n<a||n<b||n<c){
+            cout<<-1<<endl;
+            continue;
+        }
+        if(c+(n/2)<=n){
+            ans=c+(n/2);
+        }
+        else{
+            ans=c-(n/2);
+        }
+        cout<<ans<<endl;
+    }
+    return 0;
+}