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

}

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 k;
cin>>k;
int m=0;
int n=0;
int A[k];
for(int i=0;i<k;i++){
cin>>k[i];
}
for(int x=0;x<k;x+=2){
n=n+A[x];
}
for(y=1;y<k;y+=2){
m=m+A[y];
}
if(n>m){cout<<n};
else {cout<<m};
}

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 k,int n){
    if (k>=n )return 0;
    //cout << k<<endl;
    int first=Collect(A,k+2,n) +A[k];
    int sec= Collect(A,k+1,n);
    if (first > sec) return first;
    else return sec;
}
int Collectn(int A[],int k,int n, Memo& m){
    if ( m.has(k) ) return m.get(k);
    if (k>=n )return 0;
    //cout << k<<endl;
    int first=Collectn(A,k+2,n,m) +A[k];
    int sec= Collectn(A,k+1,n,m);
    if (first > sec) {m.add(k,first) ;return first;}
    else {m.add(k,sec);return sec;}
}
int main(){
    int n;
    Memo m;
    cin >>n;
    int A[n];
    for (int i=0;i<n;i++ ) cin >> A[i];
    //cout <<"INP DONE";
    cout << Collectn(A,0,n,m);

}

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

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 ms(int k, int A[], Memo&memo){
    if(memo.has(k))return memo.get(k);
    int s1=0; int s2=0;
    for(int i=0;i<k;i+=2) s1+=A[i];
    for(int i=1;i<k;i+=2) s2+=A[i];
    if(s1<s2){
        memo.add(k,s2);
        return s2;
    }
    else{
        memo.add(k,s1);
        return s1;
    }
}
int main(){
    int N; cin>>N;
    int A[N];
    Memo m;
    for(int i=0; i<N; i++){
        cin>>A[i];
    }
    int p=ms(N,A,m);
    cout<<p;
}

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 sum(int A[], int k){
    int s=0;
    for(int i=0; i<k; i++){
        s+=A[i];
    }
    return s;
}
int collect(int A[],int n, int k=0){
    if(k>n-1) return 0;
    else if(k==n-1) return A[k];
    else if(k==n-2) return A[k];
    else if((A[k] + A[k+2])>(A[k+1] + A[k+3])){
        return A[k] + collect(A,n,k+2) ;
    }
    else { return A[k+1] + collect(A,n,k+3);}
    
}

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

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