CodeGraderBench/CodeGraderBench huggingface dataset:
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include<iostream>
using namespace std;
int main(){
    int n,openCount,closedCount;
    cin>>n;
    bool isValid=true;
    for(int i=n;i>0;i--){
        char inp;
        cin>>inp;
        if(inp=='('){
            openCount+=1;
        }else{
            closedCount+=1;
        }
        if(isValid){
            if(closedCount>openCount){
                isValid=false;
                continue;
            }
        }


    }
    //loop ends,check validity condition 2
    if(closedCount!=openCount){
        isValid=false;
    }
    if(isValid){
        cout<<"VALID 0";
    }else{
        cout<<"INVALID "<<(openCount-closedCount);
    }


}

Grading criterion: Logic to check valid parenthesis
Rating A: correctly implements the logic to determine if the sequence of parentheses is valid. This includes maintaining a counter, ensuring the counter does not go negative. Student's logic could be slightly different, as long as the logic works
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include <simplecpp>
main_program{
    int n,t1=0,t2=0,t_1=0,t_2=0;char c;
    cin >> n;
    for(int i=0;i<n;i++){
        cin>>c;
        if(t1>=t2){
            if (c=='('){
                t1++;t_1++;
            }    
            else if (c==')'){
                t2++;t_2++;
            }
        }
        else{
            if (c=='(')
                t_1++;
            else if (c==')')
                t_2++;
        }
    }
    
    if(t1==t2&&t1+t2==n)
        cout<<"VALID "<<t_1-t_2;
    else 
        cout<<"INVALID "<<t_1-t_2;
}

Grading criterion: Logic to check valid parenthesis
Rating A: correctly implements the logic to determine if the sequence of parentheses is valid. This includes maintaining a counter, ensuring the counter does not go negative. Student's logic could be slightly different, as long as the logic works
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

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

int main() {
    int n; cin >> n;
    int openCount=0, closeCount=0, i=0;
    string res = "VALID";

    while (true) {
        char c; cin>>c;

        if (closeCount > openCount) res = "INVALID";

        if (c == '(') openCount++;
        else if (c == ')') closeCount++;
        
        i++;
        if (i==n) break;
    }

    if (openCount != closeCount) res = "INVALID";

    cout << res << " " << (openCount-closeCount) << endl;
}

Grading criterion: Logic to check valid parenthesis
Rating A: correctly implements the logic to determine if the sequence of parentheses is valid. This includes maintaining a counter, ensuring the counter does not go negative. Student's logic could be slightly different, as long as the logic works
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include <simplecpp>
main_program{
     int n;
      char a='(';
      char b='(';
      
      

}

Grading criterion: Logic to check valid parenthesis
Rating A: correctly implements the logic to determine if the sequence of parentheses is valid. This includes maintaining a counter, ensuring the counter does not go negative. Student's logic could be slightly different, as long as the logic works
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: C
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include<simplecpp>
main_program{
    int n;
    cin>>n;
    // char    '(';
    // char  ')';
    char x;
    int a = 0;
    int b = 0;
    while(true){
        cin>>x;
      if(x == '('){
        a = a+1;
      }
      if(x == ')'){
        b = b+1 ;
      }
      if(a+b == n){
        break;
      }
    }
    if( a == b){
        cout<<"VALID"<<" "<<"0";
    }
    if( a != b){
        cout<<"INVALID"<<" "<< (a-b);
    }
}

Grading criterion: Logic to check valid parenthesis
Rating A: correctly implements the logic to determine if the sequence of parentheses is valid. This includes maintaining a counter, ensuring the counter does not go negative. Student's logic could be slightly different, as long as the logic works
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: B
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include <simplecpp>
main_program {
        int p, openCount, closeCount ;
        cin>>p ;
        openCount = closeCount = 0 ;
        char bracket ;
        cin>>bracket ;
        bool valid=true ;
        if (bracket=='(') {
                openCount++;
                for (int i=1 ; i<p ; i++) {
                        cin>>bracket ;
                        if (bracket=='(') openCount++ ;
                        if (bracket==')') closeCount++ ;
                }
        }
        else {
                closeCount++;
                valid = false ;
                for (int i=1 ; i<p ; i++) {
                        cin>>bracket ;
                        if (bracket=='(') openCount++ ;
                        if (bracket==')') closeCount++ ;
                }
        }
        int N = (openCount - closeCount) ;
        if (N==0 && valid) cout<<"VALID " ;
        else cout<<"INVALID " ;
        cout<<N ;
}

Grading criterion: Logic to check valid parenthesis
Rating A: correctly implements the logic to determine if the sequence of parentheses is valid. This includes maintaining a counter, ensuring the counter does not go negative. Student's logic could be slightly different, as long as the logic works
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: B
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include<simplecpp>
main_program{
	int n;
	cin >> n;
	char a,b,c,d;
	cin >> a >> b >> c >> d;
	

	if(n == 4 && a==')' && b=='(' && c == ')' && d == '('){
		cout << "INVALID 0";
	}
	else if(n == 4 && a=='(' && b=='(' && c=='(' && d == ')'){
		cout << "INVALID 2";
	}
	else{
		cout << "VALID 0";
	}
}

Grading criterion: Logic to check valid parenthesis
Rating A: correctly implements the logic to determine if the sequence of parentheses is valid. This includes maintaining a counter, ensuring the counter does not go negative. Student's logic could be slightly different, as long as the logic works
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: C
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include <simplecpp>

main_program {
    int o, c;
    bool flag = true;
    o = c = 0;

    int n;
    cin >> n;

    char s;
    cin >> s;
    
    for (int i = 1; i <= n; i++, cin >> s) {
        if (s == '(') {
            o++;
        }
        else if (s == ')') {
            c++;
        }
        if (c > o) {
            flag = false;
        }      
    }
    if (!flag) {
        cout << "INVALID " << (o-c);
    }
    else {
        cout << ((o-c==0)?"VALID ":"INVALID ") << (o-c);
    }
}

Grading criterion: Logic to check valid parenthesis
Rating A: correctly implements the logic to determine if the sequence of parentheses is valid. This includes maintaining a counter, ensuring the counter does not go negative. Student's logic could be slightly different, as long as the logic works
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include <simplecpp>
main_program {
    int n , a=0 ,l=0;
    cin>>n;
    repeat(n){
        char z;
        cin>>z;
        if (z=='(') {a=a+1;}
        if (z==')') {a=a-1;}
        if (l>=0) {l=a;}
    }
    if (a!=0) {cout << "INVALID " << a;}
    if (a==0) { if (l<0) {cout << "INVALID "<< a;}
    else cout<< "VALID "<<a;}
}

Grading criterion: Logic to check valid parenthesis
Rating A: correctly implements the logic to determine if the sequence of parentheses is valid. This includes maintaining a counter, ensuring the counter does not go negative. Student's logic could be slightly different, as long as the logic works
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include <simplecpp>

main_program
{
    int n; cin >> n;
    int f=0,b=0;
    char x;
    bool val = true;
    for(int i=0; i<n; i++)
    {
        cin >> x;
        if(x == '(')
        {
            f++;
        }
        else b++;
        if( b > f ) val = false;
    }
    if(f != b) val = false;
    if(val) cout << "VALID ";
    else cout << "INVALID ";
    cout << f-b;
}

Grading criterion: Logic to check valid parenthesis
Rating A: correctly implements the logic to determine if the sequence of parentheses is valid. This includes maintaining a counter, ensuring the counter does not go negative. Student's logic could be slightly different, as long as the logic works
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include<simplecpp>
main_program{
    unsigned int n;
    cin>>n;
    int O=0,C=0;
    char first_x;
    cin>>first_x;
    if(first_x=='('){
        O+=1;
    }
    else if(first_x==')'){
        C+=1;
    }
    repeat(n-1){
        char x;
        cin>>x;
        if(x=='('){
            O+=1;
        }
        else if(x==')'){
        C+=1;
        }
    }
    int N=O-C;
    if(N!=0){
        cout<<"INVALID "<<N;
    }
    else{
        if(first_x==')'){
            cout<<"INVALID "<<N;
        }
        else if(first_x=='('){
            cout<<"VALID "<<N;
        }
    }
}

Grading criterion: Logic to check valid parenthesis
Rating A: correctly implements the logic to determine if the sequence of parentheses is valid. This includes maintaining a counter, ensuring the counter does not go negative. Student's logic could be slightly different, as long as the logic works
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include <iostream>
using std::cin; using std::cout;

int main() {
    int n, pos = 0, neg = 0;
    bool valid = true;
    cin >> n;
    for (int i = 0; i < n; i++) {
        char c;
        cin >> c;
        if (i == 0 && c == ')') {
            neg++;
            valid = false;
            continue;
        }
        if (c == '(') pos++;
        else if (c == ')') neg++;
    }
    if (pos == neg && valid) cout << "VALID " << pos-neg;
    else cout << "INVALID " << pos-neg;  
}

Grading criterion: Logic to check valid parenthesis
Rating A: correctly implements the logic to determine if the sequence of parentheses is valid. This includes maintaining a counter, ensuring the counter does not go negative. Student's logic could be slightly different, as long as the logic works
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A