What can be an efficient way to do this code, it gives a tle

#1

#include
#include
using namespace std;

int main() {
int n,a[100],i,j,k,l,sum,max,t,x;
cin>>t;
for(x=1;x<=t;x++)
{
cin>>n;
for(i=0;i<n;i++)
cin>>a[i];
max=INT_MIN;

for(i=0;i<n;i++)
{
	for(j=i;j<n;j++)
	{
		sum=0;
		for(k=i;k<=j;k++)
		sum+=a[k];
		if(sum>max)
		max=sum;
		if(j==n-1)
		{
			for(l=0;l<i;l++)
			{
				sum=0;
				for(k=i;k<=j;k++)
					sum+=a[k];
				for(k=0;k<=l;k++)
					sum+=a[k];
				if(sum>max)
					max=sum;
			}
				
		}
		
	}
}
cout<<max;
}

return 0;

}

#2

Hello @muskanrathore01_4aab76aa7dafc189,

  • Increase your array size from 100 to 1000.
  • Precompute prefix array sum which helps you to find the sum of a range in O(1)
    Efficient Code
#3

I am not able to solve it I even used the approach by using kadane’s algorithm to reduce time complexity

#4

#include #include using namespace std; int main() { int n,a[1000],i,j,k,l,sum,max,t,x,min,sum2,s1; cin>>t; for(x=1;x<=t;x++) { cin>>n; for(i=0;i<n;i++) cin>>a[i]; max=INT_MIN; min=INT_MAX; sum=0; for(i=0;i<n;i++) { sum+=a[i]; if(max<sum) max=sum; if(sum<0) sum=0; } sum2=0; for(i=0;i<n;i++) { sum2+=a[i]; if(min>sum2) min=sum2; if(sum2>0) sum2=0; } sum=0; for(i=0;i<n;i++) sum+=a[i]; s1=sum-min; if(s1==0) cout<<max; else if(s1>max) cout<<s1; else if(s1<max) cout<<max; } return 0; }

#5

@muskanrathore01_4aab76aa7dafc189,

You need to print the max sum for each test case in a new line.
Change this:

 if(s1==0)
     cout<<max;
 else if(s1>max)
     cout<<s1;
 else if(s1<max)
     cout<<max;

to:

if(s1==0)
   cout<<max<<endl;
else if(s1>max)
   cout<<s1<<endl;
else if(s1<max)
   cout<<max<<endl;
#6

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