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 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: 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 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: 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>
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: 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 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: 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 namespace std;
int Collect(int A[],int k){
 if(1>k>100) return 0;
 return Collect(A,k+2)+A[k];
}
int main() {
int k;
cin>>k;
int A[k];
cin>>A[k];
cout<<Collect(A,k)<<endl;
    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<simplecpp>
void collect(int A[],int n){
    for(int k=0;k<n;k++){
    if(k>100){
        return 0;
    }
    else {
        return max=A[k]+collect(A,k+1);
    }
    }
}
int main(){
    int n;
    cin>>n;
    int A[400];
    for(int i=0;i<n;i++){
        cin>>A[i];
    }
    maxamount=collect(A,n);
    cout<<maxamount<<endl;
    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;
int main(){
    int n;
    cin>>n;
    int arr[n];
    for (int i=0;i<n;i++){
        cin>>arr[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>
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: 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){
    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: 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 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: 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 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: 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 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: 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: B