1646B - Quality vs Quantity - CodeForces Solution


brute force constructive algorithms greedy sortings two pointers *800

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Python Code:



testcases = int(input())
numbers = []

for i in range(testcases):
    n = int(input())
    numbers.append(list(map(int,input().split())))
    
    
def checkSequence(num):
        l = 0
    r = len(num) - 1
    redCount = 0
    redSum = 0
    blueCount = 0
    blueSum = 0
    
    while l <= r:
        if redCount < blueCount and redSum > blueSum:
            return 'YES'
        if redSum <= blueSum:
            redSum += num[r] 
            redCount += 1
            r -= 1
        elif blueCount <= redCount:
            blueSum += num[l]
            blueCount += 1
            l += 1
        
    if redCount < blueCount and redSum > blueSum:
            return 'YES'
    return 'NO'
    
for num in numbers:
    num.sort()
    print(checkSequence(num))
        
        
    
    

C++ Code:

#include<bits/stdc++.h>
using namespace std;


int main()
{
 long long t,n;
 cin>>t;

 while(t--){
  cin>>n;
   long long arr[n];
  for(long long i=0;i<n;i++){
    cin>>arr[i];
  }
sort(arr,arr+n);
// reverse(arr,arr+n);
if(n%2==0){
   long long l=(n/2-1);
   long long sum=arr[0]; long long cnt = 0;
   int c=0;
    for(int i=0;i<(l);i++){
     sum=sum+arr[i+1];
     cnt=cnt+arr[(n-1)-i];
    //  cout<<cnt<<" "<<sum<<endl;
       if(cnt>sum){
        cout<<"YES"<<endl;
        c=1;
       break;
       }
    }
 if(c==0){
  cout<<"NO"<<endl;
  }
}else
{
  long long l=(n/2);
  long long sum=arr[0]; long long cnt = 0;
  int c=0;
for(int i=0;i<(l);i++){
  sum=sum+arr[i+1];
  cnt=cnt+arr[(n-1)-i];
  // cout<<cnt<<" "<<sum<<endl;
  if(cnt>sum){
    cout<<"YES"<<endl;
    c=1;
    break;
  }
}
if(c==0){
  cout<<"NO"<<endl;
}
}
 }
    return 0;
}


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