Bismillahir Rahmanir Rahim
First sort the array then take first two elements sum with a variable & take just last element as a variable
First element sum to ahead like (0+1) -> 2 -> 3
Last element sum to backward like (n-1) -> n-2 -> n-3
after adding ahead & backward check backward is high or not.If high then break & given YES.else NO.
///Author: Tanvir Ahmmad
///CSE,Islamic University,Bangladesh
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<vector>
#include<iterator>
#include <functional> ///sort(arr,arr+n,greater<int>()) for decrement of array
/*every external angle sum=360 degree
angle find using polygon hand(n) ((n-2)*180)/n*/
///Floor[Log(b) N] + 1 = the number of digits when any number is represented in base b
using namespace std;
typedef long long ll;
int main()
{
ll tst,n;
cin>>tst;
while(tst--)
{
cin>>n;
ll arr[n+4],vis[n+7],flag=0,p=6;
for(ll i=0;i<n;i++) cin>>arr[i],vis[i]=0;
sort(arr,arr+n);
ll low=arr[0]+arr[1],start=1;
ll high=arr[n-1],endu=n-1;
vis[0]=vis[1]=vis[n-1]=1;
if(low<high)
flag=1;
while(start<endu && flag==0)
{
start++,endu--;
if(vis[start]==0) low+=arr[start],vis[start]=1;
if(vis[endu]==0) high+=arr[endu],vis[endu]=1;
if(low<high)
flag=1;
}
if(flag) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}
///Alhamdulillah
Problem Link:https://codeforces.com/problemset/problem/1646/B
Comments
Post a Comment