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 main(){
int A=0;
int k=0;
int collect(A,k);
{
    if (Collect(A,k) = A[k] + Collect(A,k+2)>(A,k) = A[k] + Collect(A,k+2)>Collect(A,k) = Collect(A,k+1))
return Collect(A,k) = A[k] + Collect(A,k+2);
else 
return Collect(A,k) = Collect(A,k+1);
}
int main;
{
    int A;
    int k;
    cout<<Collect(A,k);
}
}

Grading criterion: Base condition
Rating A: Correctly handles the base conditions and returns appropriately

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 mone[n];
    for(int i=0;i<n;i++){
        cin>>mone[i];
    }
    int sum0 =0;
    int sum1=0;
    for(int j=0; j<n;j=(j+2)){
        int sum0 = sum0 + mone[j];
    }
    for(int k=1;k<n;k=(k+2)){
        int sum1 = sum1 + mone[k];
        
    }
if(sum0 >= sum1){cout<<sum0;}
else{cout<<sum1;}

return 0;

}

Grading criterion: Base condition
Rating A: Correctly handles the base conditions and returns appropriately

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 k,int n,Memo memo){
    if(memo.has(k)) return memo.get(k);
    if(k==n-1||k==n-2) {
        memo.add(k,A[k]);
        return memo.memo[k];
    }
    int value=A[k]+Collect(A,k+2,n,memo)>Collect(A,k+1,n,memo)?A[k]+Collect(A,k+2,n,memo):Collect(A,k+1,n,memo);
    memo.add(k,value);
    return memo.memo[k];
}
int main(){
    int n;
    Memo memo;
    cin>>n;
    int A[n];
    for(int i=0;i<n;i++){
        cin>>A[i];
    }
    cout<<Collect(A,0,n,memo);
}

Grading criterion: Base condition
Rating A: Correctly handles the base conditions and returns appropriately

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

Grading criterion: Base condition
Rating A: Correctly handles the base conditions and returns appropriately

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;
int memo[100], n;
int Solve(int i, int A[]) {
    //cout << i<< " " << memo[i]<<"\n";
    if (memo[i]==-1&&i<n-2)
        memo[i]=max(Solve(i+1,A),A[i]+Solve(i+2,A));
    //cout << i<< " " << memo[i]<<"\n";
    return memo[i];
}
int main() {
    int A[100];
    cin >> n;
    for (int i=0;i<n;memo[i]=-1,cin >> A[i++]);
    memo[n-1]=A[n-1];
    memo[n-2]=max(A[n-2],A[n-1]);
    cout << Solve(0,A);
    //for (int i=0;i<n;cout << memo[i++]<<endl);
}

Grading criterion: Base condition
Rating A: Correctly handles the base conditions and returns appropriately

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 mainspace std;
int Collect(int A; int k[]){
    
}
int main(){
    int N;
}

Grading criterion: Base condition
Rating A: Correctly handles the base conditions and returns appropriately

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

Grading criterion: Base condition
Rating A: Correctly handles the base conditions and returns appropriately

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;


struct Memo {

    vector<int>memo;

    int Collect(int A[], int N, int k) {
        int amount = 0;
        if(k>=N) {
            amount =0;
            return 0;
        }
        if(k==N-1) {
            amount += A[k];
            memo[k] = amount;
            return amount;
        }
        if(k<N-1) {
        
            if(memo[k]!=-1){
                amount = memo[k];
                return amount;
            }
            int including = A[k] + Collect(A,N,k+2);
            int not_including = Collect(A,N,k+1);
            if(including >=not_including) amount += including;
            else amount += not_including;
            memo[k] = amount;
            return amount;
        }
        return amount;
    }
};

int main() {
    int N;
    cin >> N;
    int A[100];
    for(int i=0; i<N; i++) {
        cin >> A[i];
    }
    Memo cm;
    vector<int>memo(100,-1);
    cm.memo = memo;
    int amount = cm.Collect(A,N,0);
    cout << amount;
}

Grading criterion: Base condition
Rating A: Correctly handles the base conditions and returns appropriately

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

Grading criterion: Base condition
Rating A: Correctly handles the base conditions and returns appropriately

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 std::cin; using std::cout; using std::endl;
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 maxamount(int amount[],int n,int k=0){int a,b;
for(;k<n;){
a=amount[k]+maxamount(amount,n,k+2);
b=maxamount(amount,n,k+1);
} ;if (a>b){return a;}else {return b;}
}
int main(){
    int n;
    cin>>n;
    int amount[n];
    for (int i=0;i<n;i++){cin>>amount[i];}
    cout<<maxamount(amount,n);
}

Grading criterion: Base condition
Rating A: Correctly handles the base conditions and returns appropriately

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 k=0){
        if(N==2) return (arr[0]>arr[1])? arr[0]:arr[1];
        if(k>=N-2) return 0;
        if(arr[k]+arr[k+2]>arr[k+1]){
            collect(arr,N,k+3);
            return arr[k]+ arr[k+2];
        }
       else if((arr[k]+arr[k+2]<arr[k+1]))
        {
            collect(arr, N,k+2);
            return  arr[k+1];
        }
        else{
            collect(arr,N,k+3);
            return arr[k+1];
        }
        

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

Grading criterion: Base condition
Rating A: Correctly handles the base conditions and returns appropriately

Rating B: Some issue. 

Rating C: Major issue. 

Rating D: 0
Assigned grade: A