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, int n, Memo& ne){
    int collect1 = 0, collect2 = 0;
    if(k + 2 < n){
        if(ne.filled[k+2]){
            collect1 = A[k] + ne.memo[k+2];
        } else {
            ne.memo[k+2] = Collect(A, k + 2, n, ne);
            ne.filled[k+2] = 1;
        }
        collect1 = A[k] + ne.memo[k+2];
    }
    else
        collect1 = A[k];
    
    if(k + 1 < n){
        collect2 = Collect(A, k+1, n, ne);
    } else collect2 = 0;
    if(collect1 > collect2){
        return collect1;
    }else {
        return collect2;
    }
}

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

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and both conditions 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>
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 n, Memo& memo) { 
    for(int i=0; i<n; i++){
        cin>>A[i];
    }
    if memo.has(n)
    return memo.get(n);
   else if( n==0) return 0;
    else if (n==1) return A[0];
  else {collect 1(A,n)  = A[n] + collect (A, n+2 );
        collect2 (A,n)= A[n]+ collect(A,n+1); 
        
        if (collect1>= collect2) ans=collect1;
        else ans= collect2;
        memo.add(ans,n);
        return ans; }

}
int main(){
    int n;
    cin>>n;
    int A[n]={};
    cout<<collect(A,n )<<endl;
}

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and both conditions 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>
using namespace std;
int Collect(int A[], int n, int k) {
    if(k==n-1) {
        return A[k];
    }
    if(k>=n-1) {
        return 0;
    }
    int include = Collect(A, n, k+2) + A[k];
    int exclude = Collect(A, n, k+1);
    return (include>exclude ? include:exclude);
}
int main() {
    int n;
    cin>>n;
    int A[n];
    for(int i=0; i<n; i++) {
        cin>>A[i];
    }
    cout << Collect(A, n, 0);
}

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and both conditions 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<array>
using namespace std;
int main(){
    int N;
    cin>> N;
    int arrayA[N];
     int k=0;
     int sum1=0;
     int sum2=0;
     for(int i=0;i<N;i++){
        cin>>arrayA[i];
     }
    for(int i=0 ; i<N; i++){
        if (i%2==0){
            sum1 +=arrayA[i];
        }
        else { sum2 +=arrayA[i];
        }
}  
if(sum1 >sum2){
    k== sum1;
}
else {
    k==sum2;
}
cout<< k<<endl;
}

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and both conditions 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; 

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 N){ 
if (N==1){return A[N];}
else if( N==2 ){ return max(A[0],A[1]);} 
}

int main() { 
int N; 

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

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and both conditions 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 A[N];
    for(int i=0;i<N;i++){
        cin>>A[i];
    }
    int s1=0,s2=0;
    for(int i=0;i<N/2;i++){
        s1+=A[2*i];
        s2+=A[2*i+1];
    }
    if(s1>s2){cout<<s1;};
    if(s2>=s1){cout<<s2;};
}

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and both conditions 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>

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

    max=A[n-2]+Collect(n-2,A);
    
    return max;
}    
    

int main() {
    int N;
    std::cin>>N;
    
    int A[N];

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

    std::cout<<Collect(N,A);
}

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and both conditions 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){
    int k=0;
    if( k>n)
    return 0;
    return a[k]+ collect(a,k+2);

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

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and both conditions 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>
int collector(A [],int k,int n){
    if(k==0) return 0;
    int sum1=0,sum2=0;
    for(int i=0;i<n;i++){
        sum1=A[i]+sum1(A[],i+2,n);
        sum2=
    }
    if(sum1>sum2) return sum2;
    else return sum2;
}
using namespace std;
int main(){
    int n;
    cin>>n;
    int A[n];
    for(int i=0;i<n;i++){
        cin>>A[i];
    }
  cout<<collector(A,k,n);
  return 0;
}

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and both conditions 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 <cmath>
#include <algorithm>
using namespace std;

int maxAmount(int arr[] , int k , int n){
    if(k>=n) return 0;
    
    return max(maxAmount(arr,k+2, n) + arr[k] , maxAmount(arr,k+1 ,n));
}

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

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and both conditions 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> 

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: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and both conditions to rob or not rob the house
Rating B: Some issue.
Rating C: Major issue. 

Rating D: 0
Assigned grade: C