CodeGraderBench/CodeGraderBench huggingface dataset:
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include <iostream> 

using namespace std;

struct Memo {

   static const int NMAX = 200;
   bool filled[NMAX] = {};
   int memo[NMAX];
   bool has(unsigned i) { return filled[i]; }
   int get(unsigned i) { return memo[i]; }
   void add(unsigned i, int val) {
       memo[i] = val;
       filled[i] = true;
   }
};

int Collect(int A[], int k) {
    if (k > ) {
        return 0;
    }
}

int main() {
    int N;
    cin >> N;
    int A[N];
    int v;
    
    for(int i  = 0; i < N; i++) {
        cin >> v;
        A[i] = v;
    }

    

}

Grading criterion: Deciding rob or not rob
Rating A: Correctly decides whether to rob or not rob the house.
Rating B: Some issue.
Rating C: Major issue. 

Rating D: 0
Assigned grade: C
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include<iostream>
using namespace std;

int Collect(int A[],int k)
{  
  if(k==0) return 0;

  int a= A[k-1]+Collect(A,k-2);
  int b= Collect(A,k-1);
  if(a>b) return a;
  else return b;

}
int main ()
{    
     int n;
     cin>>n;
     int a[n];
     for(int i=0;i<n;i++)
     {
        cin>>a[i];
     }
     cout<<Collect(a,n)<<endl;
    return 0;
}

Grading criterion: Deciding rob or not rob
Rating A: Correctly decides whether to rob or not rob the house.
Rating B: Some issue.
Rating C: Major issue. 

Rating D: 0
Assigned grade: B
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include<iostream>
using namespace std;
int collect1(int*a,int k){
    if(k=0)return a[0];
    return(a[k]+collect1(a,k-2));
}
int collect2(int*a,int k){
    if(k=1)return a[1];
    return(a[k-1]+collect2(a,k-3));
}
int collect(int*a,int n){
    return(collect1(a,n)>collect2(a,n)?collect1(a,n):collect2(a,n));
}
int main(){
    int N;
    cin>>N;
    int A[N];
    for(int i=0;i<N;i++){
        cin>>A[i];
    }
    cout<<collect(A,N-1);
}

Grading criterion: Deciding rob or not rob
Rating A: Correctly decides whether to rob or not rob the house.
Rating B: Some issue.
Rating C: Major issue. 

Rating D: 0
Assigned grade: C
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include <iostream>
using namespace std ;
int main () {

    int N ;
    cin >> N ;
    int array [N]; 

    for (int i = 0 ; i < N ; i++)
    cin >> N ;
}

Grading criterion: Deciding rob or not rob
Rating A: Correctly decides whether to rob or not rob the house.
Rating B: Some issue.
Rating C: Major issue. 

Rating D: 0
Assigned grade: C
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include<iostream>
using namespace std;

int collect(int arr[],int k,int n){
    if(k>n-3){return 0;}

    collect(arr,k,n)= arr[k];
    if(collect(arr,k,n)>=collect(arr,k+1,n)){return collect(arr,k,n);}
    else{return collect(arr,k,n);}
}


int main(){
    int n;
    cin>>n;
    int arr[n];
    for(int i=0;i<n;i++){
        cin>>arr[i];
    }
    cout<<collect(arr,0,n);
    return 0;
}

Grading criterion: Deciding rob or not rob
Rating A: Correctly decides whether to rob or not rob the house.
Rating B: Some issue.
Rating C: Major issue. 

Rating D: 0
Assigned grade: C
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include<iostream>
using namespace std;
int Collect(int arr[], int n){
    int even=0, odd=0;
    for(int i=0; i<n; i++){
        if(i%2==0){
            even += arr[i];
        } else if(i%2!=0){
            odd += arr[i];
        }
    }
    if(even>odd) {
        return even;
        } else{
            return odd;
        } 
        
}
int main(){
    int N;
    cin>>N;
    int A[N];
    for(int i=0; i<N; i++){
        cin>>A[i];
    }
    cout<<Collect(A,N)<<endl;

}

Grading criterion: Deciding rob or not rob
Rating A: Correctly decides whether to rob or not rob the house.
Rating B: Some issue.
Rating C: Major issue. 

Rating D: 0
Assigned grade: C
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include<iostream>
#include<vector>
using namespace std;

vector<int> memo;

int Collect(int A[],int k,int n)
{
    if(k>=n) return 0;
    if(memo[k]!=-1) return memo[k];
    int a=A[k]+Collect(A,k+2,n);
    int b=Collect(A,k+1,n);

    memo[k]=(a>b? a:b);
    return memo[k];
}
int main()
{
    int n;
    int A[100];
    cin>>n;
    for(int i=0;i<n;i++) cin>>A[i];
    memo.resize(n+1,-1);
    cout<<Collect(A,0,n);
}

Grading criterion: Deciding rob or not rob
Rating A: Correctly decides whether to rob or not rob the house.
Rating B: Some issue.
Rating C: Major issue. 

Rating D: 0
Assigned grade: A
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include <iostream>
#include <cmath>
using namespace std;

int f(int n, int arr[]) {
    int sum1 = 0;
    int sum2 = 0;

    for (int i = 0; i < n; i+=2) {
        sum1 += arr[i];
    }
    for (int i = 1; i < n; i+=2) {
        sum2 += arr[i];
    }

    return max(sum1, sum2);
}

int main() {
    int n; cin >> n;
    int arr[n];

    for (int i = 0; i < n; i++) {
        cin >> arr[i];
    }

    cout << f(n, arr);
}

Grading criterion: Deciding rob or not rob
Rating A: Correctly decides whether to rob or not rob the house.
Rating B: Some issue.
Rating C: Major issue. 

Rating D: 0
Assigned grade: C
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include<iostream>
using namespace std;

int collect(int a[] ,int k,int n){

if(k>n-1){return 0;}

return a[k]+collect(a,k+2,n) ;

}

int main(){
int n; cin>>n;
int a[n];
for(int i=0 ; i< n ; i++){cin>>a[i];}

int ans1=collect(a,0,n);
int ans2 = collect(a,1,n);

if(ans1>=ans2){cout<<ans1;}
else cout<<ans2;
}

Grading criterion: Deciding rob or not rob
Rating A: Correctly decides whether to rob or not rob the house.
Rating B: Some issue.
Rating C: Major issue. 

Rating D: 0
Assigned grade: C
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include<iostream>
using namespace std;
int Collect(int A[],int k)
{
    if(k==0)
    {
        return A[0];
    }
    else if(k==1)
    {
        return A[1];
    }
    return A[k] + Collect(A,k-2);
}
int main()
{
    int N,a=0,b=0;
    cin>>N;
    int A[N];
    for(int i=0;i<N;i++)
    {
        cin>>A[i];
    }
    a=Collect(A,N-1);
    b=Collect(A,N-2);
    if(a>b)
    cout<<a;
    else
    cout<<b;
}

Grading criterion: Deciding rob or not rob
Rating A: Correctly decides whether to rob or not rob the house.
Rating B: Some issue.
Rating C: Major issue. 

Rating D: 0
Assigned grade: C
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include<iostream>
using namespace std;
int Collect(int A[],int n,int k=0){
    if(k>=n) return 0;
    return A[k]+Collect(A,n,k+2)>Collect(A,n,k+1)?A[k]+Collect(A,n,k+2):Collect(A,n,k+1);
}
int main(){
    int n;
    cin>>n;
    int A[n];
    for(int i=0;i<n;i++){
        cin>>A[i];
    }
    cout<<Collect(A,n);
}

Grading criterion: Deciding rob or not rob
Rating A: Correctly decides whether to rob or not rob the house.
Rating B: Some issue.
Rating C: Major issue. 

Rating D: 0
Assigned grade: B
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include<iostream>
using namespace std;


int Collect (int a[],int k,int n){
    
if(k>n-4)
return 0;
if(a[k]+a[k+2]>a[k+1]+a[k+3])
return a[k]+Collect(a,n,k+2);
else 
return a[k+1]+Collect(a,n,k+3);
}
int main(){
int n,a[n],k=0;
 cin>>n;
for(int i=0;i<n;i++){
    cin>>a[i];
}
Collect(a,k,n);
}

Grading criterion: Deciding rob or not rob
Rating A: Correctly decides whether to rob or not rob the house.
Rating B: Some issue.
Rating C: Major issue. 

Rating D: 0
Assigned grade: C