My soln. for q1

Author: HARSH4885-svg

Problems/Mathematics/Day-01/sol/HARSHVARDHAN/.cph/.Solution1.cpp_239a8db3afbbc7d867c830a8c249ff3e.prob

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+{"name":"Local: Solution1","url":"c:\\Users\\Harshvardhan\\Downloads\\CP-Chronicles\\Problems\\Mathematics\\Day-01\\sol\\HARSHVARDHAN\\Solution1.cpp","tests":[{"id":1766761298231,"input":"7\n6 2 4\n2 3 1\n2 4 10\n5 3 4\n1 3 2\n2 5 4\n4 3 2","output":"8\n-1\n-1\n-1\n4\n1\n-1"}],"interactive":false,"memoryLimit":1024,"timeLimit":3000,"srcPath":"c:\\Users\\Harshvardhan\\Downloads\\CP-Chronicles\\Problems\\Mathematics\\Day-01\\sol\\HARSHVARDHAN\\Solution1.cpp","group":"local","local":true}

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

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+/*PROBLEM STATEMENT :
+
+There are n people standing in a circle. They are numbered from 1 to n in a clockwise order. The circle is even (i.e. n is even).
+
+Each person is looking at the person standing exactly opposite him in the circle.
+
+You are given three distinct integers a, b, and c. It is known that person a is looking at person b.
+
+Determine the number of the person that person c is looking at.
+If there are multiple answers, print any of them.
+If there is no answer, print -1.
+
+Approach : Total(i.e. n) = 2*(difference of a and b), so if any of the a,b,c is greater than total, it will return -1. Else we should return c + |a-b| or c-|a-b| (only one of them will be feasible)
+
+Time Complexity: O(1) because we have created a formula directly.
+Space Complexity : O(1) because no. of inputs is fixed 
+
+Input
+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 */
+
+#include<bits/stdc++.h>
+#include<array>
+#include<unordered_set>
+using namespace std;
+typedef long long ll;
+#define int            long long int
+#define F              first
+#define S              second
+#define pb             push_back
+#define si             set <int>
+#define vi             vector <int>
+#define pii            pair <int, int>
+#define vpi            vector <pii>
+#define vpp            vector <pair<int, pii>>
+#define mii            map <int, int>
+#define mpi            map <pii, int>
+#define spi            set <pii>
+#define endl           "\n"
+#define sz(x)          ((int) x.size())
+#define all(p)         p.begin(), p.end()
+#define double         long double
+#define que_max        priority_queue <int>
+#define que_min        priority_queue <int, vi, greater<int>>
+#define bug(...)       __f (#__VA_ARGS__, __VA_ARGS__)
+#define vin(a) for(int i = 0; i< a.size();i++) cin>>a[i]
+#define vout(a) {for(auto x : a) cout << x << " "; cout << endl;}
+#define pp(a) for(auto x : a) cout << x.F << " " << x.S << endl
+#define print(a,x,y) {for(int i = x; i < y; i++) cout<< a[i]<< " "; cout << endl}
+
+inline int power(int a, int b){int x = 1;while (b){if (b & 1) x *= a;a *= a;b >>= 1;}return x;}
+template <typename Arg1>
+void __f (const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << endl; }
+template <typename Arg1, typename... Args>
+void __f (const char* names, Arg1&& arg1, Args&&... args)
+{
+    const char* comma = strchr (names + 1, ',');
+    cout.write (names, comma - names) << " : " << arg1 << " | "; __f (comma + 1, args...);
+}
+
+bool isPrime(int n){ if(n <= 1) return false; for(int i=2;i*i<=n;i++) if(n % i == 0) return false; return true; }
+ll gcd(ll a, ll b){ return b ? gcd(b, a % b) : a; }
+ll lcm(ll a, ll b){ return a / gcd(a,b) * b; }
+ll modpow(ll a, ll b, ll m){ ll r=1; while(b){ if(b&1) r=r*a%m; a=a*a%m; b>>=1;} return r; }
+
+bool getBit(ll x, int k){ return (x >> k) & 1; }
+ll setBit(ll x, int k){ return x | (1LL << k); }
+ll clearBit(ll x, int k){ return x & ~(1LL << k); }
+ll toggleBit(ll x, int k){ return x ^ (1LL << k); }
+int countBits(ll x){ return __builtin_popcountll(x); }
+bool isPowerOfTwo(ll x){ return x && !(x & (x - 1)); }
+
+vector<int> sieve(int n){ vector<int> p(n+1,1), primes; p[0]=p[1]=0; for(int i=2;i*i<=n;i++) if(p[i]) for(int j=i*i;j<=n;j+=i) p[j]=0; for(int i=2;i<=n;i++) if(p[i]) primes.pb(i); return primes; }
+
+const int N = 200005;
+
+void solve() {
+    int a,b,c;
+    cin >> a >> b >> c;
+
+    int total = 2*abs(a-b);
+    if (max(max(a,b),c)>total)
+    {
+      cout << -1 << endl;return;
+    }
+    else
+    {
+      int ans = ((c-abs(a-b))>0 )? c-abs(a-b) : c+ abs(a-b) ;
+      cout << ans << endl;return;
+    }
+}
+
+int32_t main()
+{
+    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
+    int t = 1;
+    cin >> t;
+    while (t--) solve();
+    return 0;
+}