CodeGraderBench/CodeGraderBench huggingface dataset:
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include<iostream>
using namespace std;
int n;
int length=n; 
for(int i=0;i<n;i++){
    cin>>arr[i];
}
int totalpaths(int n,int arr[p],length){
       if(n==0||length<0) {return 0;}
       else if(length==0) {return 1;}
       else totalpaths=totalpaths(n,arr[p],length-arr[p])+totalBoarding(n,arr[p-1],length);

    }



int main(){
    int n;
    cin>>n;
    int length=n;
    int p=n
    totalpaths(n,arr[p],length);
}

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 issiue
Assigned grade: D
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
# include <iostream>
using namespace std;
struct Memo {
   static const int NMAX = 100;
   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 memoPath(unsigned n,int jumps[],Memo& memo){
    if(memo.has(n)) return memo.get(n);
    int ans;
    if(n<=1) ans=n;
    for(int i=n-2;n-i-jumps[i]>=0&&n>1;i--){
        ans+=memoPath(i,jumps,memo);
    } 

    memo.add(n,ans);
    return ans;

}

int main(){
    Memo memo;

    int n;
    cin>>n;

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

    }
    cout<< memoPath(n,jumps,memo);
}

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 issiue
Assigned grade: A
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include<iostream>
using namespace std;
int func(int i, int n, int jumps[]){
   int count=0,sum=0;
    if(i==n-2) return 1;
    if(i==n-1) return 0;
    for(int p=i+1; p<=i+jumps[i]; p++){
        func(p,n,jumps);
        count++;
        sum=sum+count;
    }
    return sum+1;

}
struct Memo {
   static const int NMAX = 100;
   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;
cin>>n;
int jumps[n];
for(int i=0;i<n;i++){
    cin>>jumps[i];
}
cout<<func(0,n,jumps);
}

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 issiue
Assigned grade: C
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include <iostream>
using namespace std;
int totalways(int n, int jumps[], int curr_index);
int main()
{
    int n; cin>>n; int jumps[n];
    for(int i = 0; i< n; i++)
    {
        cin >> jumps[i];
        
    }
    cout << totalways(n, jumps, 0);
}

int totalways(int n, int jumps[], int curr_index)
{
    int total = 0;
    if (n==2) return 1;
    if (curr_index > n-1) return 0;
    else if (curr_index == n-1) return 1;
    else {
    for(int i = 1;i<= jumps[curr_index] ;i++)
    {
        total+=totalways(n , jumps, curr_index + i);
    }}
    

    return total;


}

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 issiue
Assigned grade: C
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include<iostream>
using namespace std;
int now(int arr[],int i, int n){
    if(i>=n) return 0;
    if(i==n-1) return 1;
    int z=0;
    for (int x=0; x<arr[i]; x++){
        z=z+now(arr,i+x+1,n)
    }
    return z;
}
int main(){
    int n;
    cin >> n;
    int jumps[n];
    for (int i=0; i<n; i++){
        cin >> jumps[i];
    }
    int i=now(jumps,0,n);
    cout << 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 issiue
Assigned grade: B
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include<iostream>
#include<vector>
using namespace std;
int now(int n,vector<int> jump, int p){
    int static step;
    if(p== n){
        return 1;
    }
    
    step+=now(n,jump,p+1)+now(n,jump,p+2)+1;
    return step;
}

int main(){
    int n;
    cin>>n;
    vector<int> jumps(n);
    for(int i=0;i<n;i++){
        cin>>jumps[i];
    }
    cout<<now(n,jumps,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 issiue
Assigned grade: B
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include<iostream>
using namespace std;
int main()  {
    int n;
    cin >> n;
    int jumps[n];
    for(int i=0; i<n; i++){
        cin >> jumps[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 issiue
Assigned grade: D
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include <iostream>
using namespace std;

int JUMPER(int i,int n,int jumps[i])
{	
	int sum=0;
	while(i<n)
	{
		if (i+jumps[i]<n)
		{
			i=i+jumps[i];
			sum++;
			JUMPER(i,n,jumps[i]-1);
		}
        else
        {
        	JUMPER(i,n,jumps[i]-1);
        }
	}
	return sum;
}

int main()
{
	int N,k;
	cin >> N;
	int jumps[N];
	for(k=0;k<N;k++)
	{
		cin >> jumps[k];
	}
    k=0;
	JUMPER(k,N,jumps[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 issiue
Assigned grade: C
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include<iostream>
using namespace std;
int main() {
    int n;
    cin>>n;
   array jumps[n]
}

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 issiue
Assigned grade: D
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include<iostream>
using namespace std;
void noofpaths(int jumps[],n){
    int count=0;
    for(int i=0;i<n;i=i+j){
    for(int j=1;j<=i+jumps[i];j++){
        if(i=n-1)++count;
     return noofpaths(jumps,n-i);   
    }
    
    }
    cout<<count<<endl;
}
int main(){
    int N=100;
    int n;
    int jumps[N];
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>jumps[i];
    }
    noofpaths(jumps[],n)
    

}

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 issiue
Assigned grade: D
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include<iostream>
using namespace std;
int nways=0; //total ways
void number(int n, int jumps[]){
    for(int i = 1; i<=jumps[0]; ++i){
        if(i<n)  {number(n-i, &jumps[i]);} // if not reached end, check all possible paths
        else if(i==n) ++nways; // when reached end, paths found increases by one
    }
}
int main(){
    int n;cin>>n;int jumps[n];
    for(int i = 0; i< n; ++i) cin>>jumps[i];
    number(n, jumps);
    cout<<nways;
}

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 issiue
Assigned grade: C