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 std::cout; using std::cin;



int main()
{
char c; bool valid=1;
int num,openN{},closedN{}; cin>>num;

for (int i{}; i<num; i++) {
    cin>>c;
    if (c=='(') openN++;
    else if (c==')') closedN++;
    if (openN-closedN<0) valid=0;
}
cout<<(valid && openN==closedN? "VALID" : "INVALID") << ' ' << openN-closedN;
}

Grading criterion: Loop to iterate from 1 to n
Rating A: correctly iterates the loop to accept all the n characters
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{
    cin>>noskipws;
    char exp;
    int str=0,end=0,diff;
    cin>>exp;
    while(true){
        for(cin>>exp;diff>=0;cin>>exp){
            if(exp=='('){
                str+=1;
            }else if(exp==')'){
                end+=1;
            }
            diff=str-end;
        }
        if(diff<0){
            for(cin>>exp;diff>=0;cin>>exp){
            if(exp=='('){
                str+=1;
            }else if(exp==')'){
                end+=1;
            }
            diff=str-end;
            cout<<"INVALID"<<diff<<endl;
            break;
        

        }
        if(diff==0){
            cout<<"VALID"<<diff<<endl;
        }
        
    }
    
    }
}

Grading criterion: Loop to iterate from 1 to n
Rating A: correctly iterates the loop to accept all the n characters
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,c1,c2,c3,c;
    c1=0;
    c2=0;
    c3=0;
    cin >> n;
    repeat (n) {
        cin >> c;
        if ( c == '(') {
            c1 += 1;
        }
        if (c3>= 0){
            c3+=1;
        }
    }
        if (c == ')') {
        c2 += 1;
         if (c3>=0){
            c3-=1;
    }
    }

        int a=c1 -c2;
        if (c3 ==0) {
            cout << "VALID"<<" ";
        }
        else {
            cout << "INVALID"<< " ";
        }
        cout << a;
        
    }

Grading criterion: Loop to iterate from 1 to n
Rating A: correctly iterates the loop to accept all the n characters
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>

void sequence(int n, char y) {
    int n;

}



int main() {
    int n;
    cin>>n;
    char y;
    cin>>y;
    int N;
    
     

     cout<<"VALID"<<N;

}

Grading criterion: Loop to iterate from 1 to n
Rating A: correctly iterates the loop to accept all the n characters
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{
    cin>>noskipws;
    char c;
    cin>>c;
    while(c){
        cout<<c
    }
}

Grading criterion: Loop to iterate from 1 to n
Rating A: correctly iterates the loop to accept all the n characters
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>

int main(){
   int n,N,o=0,c=0;
   char a;
   cin>>n>>a;
   repeat(n){
    if(a=='('){
        o++;
    }
    else if(a==')')
    c++;
   }
    N=o-c;
    cout<<N;
}

Grading criterion: Loop to iterate from 1 to n
Rating A: correctly iterates the loop to accept all the n characters
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>

int main() {
        int n;
        cin >> n;
        int co = 0;
        int cc = 0;
        bool valid = true;

        for (int i = 0; i < n; i++) {
                char c;
                cin >> c;

                if (c == '(') {
                        co = co + 1;
                }
                else if (c == ')') {
                        cc = cc + 1;
                }

                if (cc > co) {
                        valid = false;
                }
        }

        if (cc != co) {
                valid = false;
        }

        if (valid == false) {
                cout << "INVALID " << (co - cc) << endl;
        }

        else
                cout << "VALID 0" << endl;
}

Grading criterion: Loop to iterate from 1 to n
Rating A: correctly iterates the loop to accept all the n characters
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,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: Loop to iterate from 1 to n
Rating A: correctly iterates the loop to accept all the n characters
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: Loop to iterate from 1 to n
Rating A: correctly iterates the loop to accept all the n characters
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: Loop to iterate from 1 to n
Rating A: correctly iterates the loop to accept all the n characters
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;
      char a='(';
      char b='(';
      
      

}

Grading criterion: Loop to iterate from 1 to n
Rating A: correctly iterates the loop to accept all the n characters
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: Loop to iterate from 1 to n
Rating A: correctly iterates the loop to accept all the n characters
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;
    int a,b;
    cin >> n;
    for (int t=1 ; t<=n ; t++) {
    char c; 
    cin >> c;
    if (int(c)=40) a++;
    if (int(c)=41) b++;
    
}
if ((a+b)%2=0 && a==b) cout << "VALID";
else cout << "INVALID"; 

if (a==b) cout << "0";
else if (a!=b) cout << a-b;

}

Grading criterion: Loop to iterate from 1 to n
Rating A: correctly iterates the loop to accept all the n characters
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 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: Loop to iterate from 1 to n
Rating A: correctly iterates the loop to accept all the n characters
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A