Merge pull request #422 from suzzzal/hhh

day06sol01

Author: Dheeraj-06

Problems/Binary-Search/Day-06/suzzzal/day06sol01.cpp

@@ -0,0 +1,38 @@
+// submission - https://codeforces.com/contest/2014/submission/356012129
+
+/*
+Approach:
+Binary search the extra gold x.
+For each x, compute new average and count people with wealth < half of it.
+If count > n/2, x is valid. Find minimum such x.
+
+Complexity:
+O(n log n)
+*/
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int main(){
+    int t;cin>>t;
+    while(t--){
+        int n;cin>>n;
+        vector<long long>a(n);
+        long long sum=0;
+        for(int i=0;i<n;i++){cin>>a[i];sum+=a[i];}
+        if(n<=2){cout<<-1<<"\n";continue;}
+        sort(a.begin(),a.end());
+        long long l=0,r=1e18,ans=-1;
+        while(l<=r){
+            long long mid=(l+r)/2;
+            long double avg=(long double)(sum+mid)/n;
+            long double half=avg/2.0;
+            int cnt=0;
+            for(long long v:a) if((long double)v<half) cnt++;
+            if(cnt>n/2){ans=mid;r=mid-1;}
+            else l=mid+1;
+        }
+        cout<<ans<<"\n";
+    }
+    return 0;
+}