Merge pull request #217 from ronakgoyal1/main

Solution to Day 1 problem 1

Author: Dheeraj-06

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

@@ -0,0 +1,73 @@
+/*
+Problem Statement:
+Some number of people (this number is even) are standing evenly in a circle.
+Each person looks directly at the person opposite to them.
+Given that person a is looking at person b, determine whom person c
+is looking at. If no such circle configuration exists, output -1.
+
+Approach:
+1. The distance between opposites (diff) is half the total population.
+   N = 2 * |a - b|
+2. Validity Check: a, b, and c must all be less than or equal to N.
+3. To find the opposite of c:
+   - If c <= N/2, its opposite is c + N/2.
+   - If c > N/2, its opposite is c - N/2.
+
+Time Complexity:
+O(1) per test case
+
+Space Complexity:
+O(1)
+Example:
+7
+6 2 4
+2 3 1
+2 4 10
+5 3 4
+1 3 2
+2 5 4
+4 3 2
+Output:
+8
+-1
+-1
+-1
+4
+1
+-1
+
+Submission Link : https://codeforces.com/contest/1560/submission/355273828
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    int t;
+    cin >> t;
+
+    while (t--) {
+        long long a, b, c;
+        cin >> a >> b >> c;
+
+        long long diff = llabs(a - b);
+        long long N = 2 * diff;
+
+        if (a > N || b > N || c > N) {
+            cout << -1 << endl;
+        } else {
+            long long half = N / 2;
+            long long d;
+            if (c <= half) {
+                d = c + half;
+            } else {
+                d = c - half;
+            }
+            cout << d << endl;
+        }
+    }
+
+    return 0;
+}
+
+