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>
#include <vector>
using std::cin; using std::cout; using std::vector;

int max(int a, int b) {
    return ((a > b) ? a : b);
}

int Collect(int A[], int N, vector<int>& memo, int k = 0){
    if(k >= N) return 0;
    if(memo[k] != -1) return memo[k];
    memo[k] = max(A[k] + Collect(A, N, memo, k+2), Collect(A, N, memo, k+1));
    return memo[k];
}

int main(){
    int N; cin >> N;
    int A[N];
    vector<int> memo(N, -1);
    for(int i = 0; i < N; i++) cin >> A[i];
    cout << Collect(A, N, memo);
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)

Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls.

Rating B: Some issue with memoization.
Rating C: Memoization not used

Rating D: Major issue. 

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>
using namespace std;
int main (){
    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 collector{A{}}
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)

Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls.

Rating B: Some issue with memoization.
Rating C: Memoization not used

Rating D: Major issue. 

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 money[n];
    for(int i=0;i<n;i++){
        cin>>money[i];
    }
if(n%2==0){
    int s1=0;
    for(int i=0;i<n/2;i++){
        s1+=money[2*i];
    }
    int s2=0;
    for(int i=1;i<=n/2;i++){
        s2+=money[2*i-1];
    }
    if(s1>s2)
    cout<<s1;
    else cout<<s2;
}

if(n%2!=0){
    int s1=0;
    for(int i=0;i<=n/2;i++){
        s1+=money[2*i];
    }
    int s2=0;
    for(int i=1;i<=n/2;i++){
        s2+=money[2*i-1];
    }
    if(s1>s2)
    cout<<s1;
    else cout<<s2;
}
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)

Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls.

Rating B: Some issue with memoization.
Rating C: Memoization not used

Rating D: Major issue. 

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 std::cin; using std::cout;
using std::endl;

int Collect (int A[] , int k, int N ) {
    if (k == N-2 || k == N-1) return Collect(A, k-2 ,N) + A[k];
    return std::max((A[k] + Collect(A,k+2, N)), (Collect(A,k+1,N)));
}

int main() {
    int N; cin >> N;
    int A[N];
    for (int i=0; i<N;i++) {cin >> A[i];}
    bool robbed[N] = {};
    int a = Collect(A , 0 , N); int b = Collect(A , 1 , N); 
    a>=b ? cout << a : cout << b;
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)

Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls.

Rating B: Some issue with memoization.
Rating C: Memoization not used

Rating D: Major issue. 

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,const int n){
    if(k>=n){
        return 0;
    }
    if(n==1){
        return A[0];
    }
    return A[k] + collect(int A[],int k+2,const int n);
    }
int main(){
    int n;
    cin>>n;
    int A[n];
    for(int i=0;i<n;i++){
        cin>> A[i];
    }
    
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)

Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls.

Rating B: Some issue with memoization.
Rating C: Memoization not used

Rating D: Major issue. 

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;



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, Memo& memo, int size){
    if(memo.has(k)) return memo.get(k);
    if(k >= size) return 0; 
    //base condition;
    int ans;
    int option1 = 0;
    int option2 = 0;
    option1 = Collect( A,k+1, memo,size);
    option2 = Collect(A, k+2, memo,size) + A[k];

     
     ans = (option1 > option2)? option1:option2;
     memo.add(k,ans);
     return ans;
      
    

}

int main(){

 int size ;
 cin >> size;
 int amount[size];
 for(int i=0; i<size;i++) cin>>amount[i];
 Memo memo;
 cout << Collect(amount , 0,memo,size);


}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)

Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls.

Rating B: Some issue with memoization.
Rating C: Memoization not used

Rating D: Major issue. 

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>
using namespace std;
int Collect1(int A[],int N){
    return A[N] + Collect1(A,N+2);
}
int Collect2(int A[],int N){
    return Collect2(A,N+1);
}
int main(){
    int N,a,b;
    cin>>N;
    int A[100]={};
    for(int i=0;i<N;i++){
        cin>>A[i];
    }
    a=Collect1(A,N);
    b=Collect2(A,N);
    if (a>b){
        cout<< a;
    }
    else {
        cout<< b;
    }
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)

Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls.

Rating B: Some issue with memoization.
Rating C: Memoization not used

Rating D: Major issue. 

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 money[n];
    for(int i=0;i<n;i++){
        cin>>money[i];
    }
    int sum1=0;
    int sum2=0;
    for(int i=0;i<n;i=i+2){
        sum1+=money[i];
    }
    for(int i=1;i<n;i=i+2){
        sum2+=money[i];
    }
    if(sum1<sum2){
        cout<<sum2;
    }
    else cout<<sum1;
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)

Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls.

Rating B: Some issue with memoization.
Rating C: Memoization not used

Rating D: Major issue. 

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 A[100];

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

    
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)

Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls.

Rating B: Some issue with memoization.
Rating C: Memoization not used

Rating D: Major issue. 

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

}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)

Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls.

Rating B: Some issue with memoization.
Rating C: Memoization not used

Rating D: Major issue. 

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 A[N];
    int i;
    for(i=0; i<N; ++i)
    {
        cin>>A[i];
    }
    int sum=0;
    for(i=0; i<(N/2); ++i)
    {
        sum+=A[i+2];
    }
    cout<<sum;
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)

Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls.

Rating B: Some issue with memoization.
Rating C: Memoization not used

Rating D: Major issue. 

Assigned grade: C