added Q1 solution.

Author: d2-e2-v2

Problems/Mathematics/Day-01/sol/Abhay/Q1sol.cpp

@@ -0,0 +1,52 @@
+#include <bits/stdc++.h>
+using namespace std;
+/*
+Logic:
+The main key in solving the problem is that a,b will be enought to know that the size of the area.
+Another observation would be that 1 is always facing the element  more than the element which faces 
+at the last element or the size of the circle.
+Once, you get the size of the array, we just have to check if the any of the elements are larger than 
+the size of the array. 
+On deeper checks like- a=3, b=2,c=1;
+we can observe how this isn't true but how does the conditon fulfill the case?
+--> the max check covers all small checks, like the circle don being possible
+in cases like a=1,b=6, c=100
+--> the circle condition is also checked by the max case. 
+
+So, after the checking, we need to see the answer, that can be easily given as the
+difference is same for all pairs of elements  modular nature of the circle( cool words that make sense)
+ 
+Time complexity-->
+
+The time complexity is O(1)
+The space complexity is O(1)
+
+*/
+//Soln->
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+    cin >> t;
+
+    while (t--) {
+        int a,b,c;
+        cin>>a;
+        cin>>b;
+        cin>>c;
+        int x=abs(b-a);
+    if(max({a,b,c})>(2*x))
+    cout<<-1<<endl;
+    else
+    {
+        if(c>x)
+        cout<<c-x<<endl;
+        else
+        cout<<c+x<<endl;
+    }
+    
+    }
+    return 0;
+}
+