# KETEKI2B editorial

Setter: Abhilash

Tester: Harsh Raj

Editorialist: Harsh Raj

# DIFFICULTY:

``````Easy
``````

# PREREQUISITES:

``````Partial Sum, Kadane's Algorithm
``````

# EXPLAINATION:

``````Store partial sum of the input array. Traverse the array,and
for each subarray of size k, get the number of chips. For each
iteration, update the answer if number of chips in the
``````

# TIME COMPLEXITY:

``````O(N)
``````

# SOLUTIONS:

Settler's Solution

``````	#include<bits/stdc++.h>
#define ll long long
using namespace std;
main()
{
int n,k;
cin>>n>>k;
ll a[n];
for(int i=0;i<n;i++)
cin>>a[i];
ll max_chips=0;
for(int i=0;i<k;i++)
max_chips+=a[i];
ll chips_in_hand=max_chips;
for(int i=k;i<n;i++){
chips_in_hand+=(a[i]-a[i-k]);
max_chips=max(max_chips,chips_in_hand);
}
cout<<max_chips<<"\n";
}
``````
Tester's Solution

``````	#include <bits/stdc++.h>
using namespace std;
#define ll long long int

int main() {
ll i,j,k,sum=0,n,m;
cin>>n>>k;
vector<ll> a(n);

for(i=0;i<n;i++)
cin>>a[i];
if(n < k){	//if there aren't k chips
cout<<sum<<endl;
return 0;
}

for(i=1;i<n;i++)
a[i]+=a[i-1];//storing partial sum
sum=a[k-1];
for(int index=k;index<n;index++)
sum=max(sum,a[index]-a[index-k]);

cout<<sum<<endl;

return 0;
}
``````