Merge pull request #429 from capricemoto/main

d6q1

Author: zonedoutdg

Problems/Binary-Search/Day-06/Samarth/sol1.cpp

@@ -0,0 +1,75 @@
+#include <bits/stdc++.h>
+#define endl '\n'
+typedef long long ll;
+
+using namespace std;
+
+/*
+PROBLEM STATEMENT:
+There are n people living in the town. Just now, the wealth of the i-th person was ai
+
+gold. But guess what? The richest person has found an extra pot of gold!
+
+More formally, find an aj=max(a1,a2,…,an)
+, change aj to aj+x, where x
+
+is a non-negative integer number of gold found in the pot. If there are multiple maxima, it can be any one of them.
+
+A person is unhappy if their wealth is strictly less than half of the average wealth∗
+
+.
+
+If strictly more than half of the total population n
+
+are unhappy, Robin Hood will appear by popular demand.
+
+Determine the minimum value of x
+for Robin Hood to appear, or output −1 if it is impossible.
+
+
+APPROACH:
+just following the conditions of the ques after sorting the array we get
+that x>2*n*a[mid]- sum where mid=n/2 and sum is sum of elements of array;
+the reason to sort them was that if the largest element satisfied the condition then all the subsequent
+smaller will do so automatically;
+and if n==1||n==2 ans would -1 reason simple most of them will be richer
+and if x is less than zero ans zero.
+
+
+TIME COMPLEXITY:O(n +nlogn(for sorting))so O(nlogn);
+SPACE COMPLEXITY:O(n);
+
+
+
+SUBMISSION LINK:https://codeforces.com/contest/2014/submission/356057548
+
+
+*/
+
+
+
+
+
+ll solve(){
+	int n;cin>>n;int a[n];ll sum=0;
+	for(int i=0;i<n;i++){
+		cin>>a[i];sum+=a[i];
+	}
+	if(n==1||n==2)return -1;
+	else{
+		sort(a,a+n);
+		int mid=n/2;
+		ll ans=(ll)2*n*a[mid]-sum+1;
+		if(ans>0)return ans;
+		else return 0;
+	}
+}
+
+int main(){
+    int t;cin>>t;
+    while(t--){
+    	cout<<solve()<<endl;
+    }
+
+	
+}