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;
int max(int money[],int &bag,int house,int n){
    if(house>n)
        return bag;
    bag+=money[house];
    return max(money,bag,house+2,n);
    
}

int main(){
    int n;
    cin>>n;
    int house[n];
    for(int i=0;i<n;i++){
        cin>>house[i];
    }
    int bag=0;
    int sums[2];
    sums[0]=max(house,bag,0,n);
    sums[1]=max(house,bag,1,n);
    cout<<(sums[0]>sums[1]?sums[0]:sums[1]);
}

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>
#include<vector>
using namespace std;
void coll(int arr[],int mem[],int ind,int n){
    if(ind>=n){
        return;
    }
    if(ind==n-1){
        mem[ind]=arr[ind];
        return ;
    }
    if(ind+2<n){ 
    if(mem[ind+2]==0){
    coll(arr,mem,ind+2,n);
    }
    }
    if(ind+1<n){
        if(mem[ind]==0){
    coll(arr,mem,ind+1,n);
        }
    }
    if(ind+2<n){
    mem[ind]=max(mem[ind+1],arr[ind]+mem[ind+2]);
    }else{
        mem[ind]=max(mem[ind+1],arr[ind]);
    }
}
int main(){
    int n;
    cin >> n;
    int arr[n],mem[n]={};
    for(int i=0;i<n;i++){
        cin >> arr[i];
    }
    coll(arr,mem,0,n);
    /*for(int i=0;i<n;i++){
        cout << mem[i]<<' ';
    }*/
    cout<<mem[0];
}

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

int maxmoney(int n,int arr[]) {

    if(n==0) {
        return 0;
    }

    if(n<0) {
        return 0;
    }
    



    return max(maxmoney(n-2,arr+2)+arr[0],maxmoney(n-1,arr+1));


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

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 n,int k)
{   
    int a=0;
    int b=0;
    if(n==1)
        return A[0];
    if(n==2)
    {
        if(A[1]>A[0])
            return A[1];
        else 
            return A[0];
    }
    if(k == n-3)
    {   
        if(A[k] + A[n-1] > A[n-2])
            return A[k] + A[n-1];
        else 
            return A[n-2];
    }
    if(k == n-2)
    {
        if(A[n-2]>A[n-1])
            return A[n-2];
        else 
            return A[n-1];
    }
    a = A[k] + collect(A,n,k+2);
    b = collect(A,n,k+1);
    if(a>b)
        return a;
    else 
        return b;

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

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 arr[] , int k){
        
       Memo memo;

     if(k<=0) return 0;
     
    else if(memo.has(arr[k-1])) return memo.get(arr[k-1]);

     int ans=0;

     ans = arr[k-1] + collect(arr , k-2);   

      memo.add(k-1, ans);

     

}


int main(){
   int N;
   cin>>N;
   int arr[N];
   for(int i=0; i<N ; i++){
    cin>>arr[i];
   }
   
   int ans1, ans;
   ans=collect(arr , N);
   ans1= collect(arr, N-1);
   
   cout<<(ans > ans1 ? ans: ans1);

}

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: 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 ){


}
int main(){
    int n;
    cin>>n;
    int a[n],sum=0;
   for(int i=0;i<n;++i){
    cin>>a[i];}
    for(int i=0;i<n;i+=2){
        sum+=a[i];
    }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: D
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 first( int arr[],int n,int i){
    if(i==n-1){return arr[n-1];}
    else if(i==n-2){return arr[n-2];}
    
    else if(i<n-2){
    return  arr[i] + first(arr,n,i+2);}
}
int main(){
    int n; cin>>n;
    int a[n];
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    int t=first(a,n,0); 
    int d=first(a,n,1); 
     cout<<(t>d ? t:d);
}

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

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>
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){
   return collect(A,k) = A[k] + collect(A,k+2);
   return collect(A,K)=collect(A,k+1);

}


};
int main(){
    int N;
    cin>>N;
    int A[N];
    for(int i=0;i<N;i++){
        cin>>A[i];
    }
    collect(A,k);
    cout<<"17242";

}

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 s ,int A[],int k) {
    
    if(k>s) {
        
        return 0;
    }
    
    
        int max1 = A[0] + Collect(s,A,k+2);
    int max2= Collect(s,A,k+1);
    if(max1>max2)
    return max1;
    else
    return max2;
    
    
    
    
}
int main() {
    int n;
    cin >> n;
    int A[n];
    for(int i=0;i<n;i++) {
        cin >> A[i];
    }
    int k;
    cin >> k;
    cout << Collect(n,A,k);
    
}

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