Create day006sol01.cpp

Author: Naman2251

Problems/Binary-Search/Day-06/sol/Naman2251/day006sol01.cpp

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+// Submission Link: https://codeforces.com/contest/2014/submission/356013624
+/*
+It is only not possible to call robinhood if n is 2 or 1.
+avg=(sum+k)/n.
+Since we need strictly more than half of the total population strictly less than half of the average wealth
+therefore we would just make our avg equal to a[n/2]+1 as then more than hal would be unhappy.
+(sum+k)/(n*2)=a[n/2]+1
+On solving, k=a[n/2]*n*2+1-sum
+If k<0 then no need of any coin so k=0, therefore final ans is max(k, 0).
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+ 
+int main() {
+    long long t;
+    cin>>t;
+    while(t--) {
+        long long n, sum=0;
+        cin>>n;
+        vector <long long> a(n);
+        for(int i=0; i<n; i++) {
+            cin>>a[i];
+            sum+=a[i];
+        }
+        if(n<=2) {
+            cout<<-1<<endl;
+            continue;
+        }
+        sort(a.begin(), a.end());
+        long long k= a[n/2]*n*2+1-sum;
+        cout<<max(k, 0LL)<<endl;
+    }
+}