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<simplecpp>
main_program{
    int n,N;
    int temp=0;
    char c;
    char prevc;
    int openCount=0; // number of (
    int closeCount=0; // number of )
    cin>>n;
    for(int i=1; i<=n; i++){
        cin>>c;
        if(c=='('){
            if(i==n){temp=1;} //cannot end with (
            ++openCount;
        }
        else if(c==')'){
            if(i==1){temp=1;} // cannot start with )
            ++closeCount;
        }
    }
    N = (openCount - closeCount); // difference between number of '(' and ')'
    if(N!=0){
        cout<<"INVALID"<<" "<<N;
    }
    if(N==0){
        if(temp==1){ // cannot start with')' or end with '('
            cout<<"INVALID"<<" "<<N;
        }
        else cout<<"VALID"<<" "<<N;
    }
}

Grading criterion: Printing Valid and Invalid
Rating A: correctly prints VALID / INVALID and the count based on the condition
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 a = '(' , b = ')' ,p=0,q=0;
    repeat(n){
        char c;
        cin>>c;
         if(a=='c'){p=p+1;}
         if (b=='c'){q=q+1;}

    }
    if ( p==q ){
        cout<<"valid 0"<<endl;
    }
    else {int x;x=p-q;
        cout<<"invalid "<<x<<endl;
    }

}

Grading criterion: Printing Valid and Invalid
Rating A: correctly prints VALID / INVALID and the count based on the condition
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 a=0;
        int b=0;
                char c;
        for(int i=0;i<n;i++){
                cin>>c;
                if(c=='{'||c=='(') a++;
                if(c=='}'||c==')') b++;
        }
        if(((a-b)==0)&&(c==')'||c=='}')) cout<<"VALID " <<a-b;
        else cout<<"INVALID "<<a-b;}

Grading criterion: Printing Valid and Invalid
Rating A: correctly prints VALID / INVALID and the count based on the condition
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;
char x;
int m=0;
bool z,y;
cin>>n;
cin>>x;
if(x==')'){y=true;m=m-1;}
if(x=='('){z=true;m=m+1;}
if(z)
{
repeat(n-1)
    {
    cin>>x;
    if(x==')'){m=m-1;}
    if(x=='('){m=m+1;}
    }   
}
if(y)
{repeat(n-1)
    {
    cin>>x;
    if(x==')')m=m-1;
    if(x=='(')m=m+1;
    }}
if(y)cout<<"INVALID "<<m;
if((z)&&(m==0))cout<<"VALID "<<m;
if((z)&&(m!=0))cout<<"INVALID "<<m;
    return 0;
}

Grading criterion: Printing Valid and Invalid
Rating A: correctly prints VALID / INVALID and the count based on the condition
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 input;
    cin>>input;
    if(0<input-'0'<9)
    {
        cin<<input;
    }
    else{
    if(input=='(')
    {
        i=i+1;
    }
    else
    {i=i-1;}
    
    }
    cout<<"Valid "<<i;
}

Grading criterion: Printing Valid and Invalid
Rating A: correctly prints VALID / INVALID and the count based on the condition
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;
    int n;
    cin >> n;
    int open,close;
    char openclose;
    for(cin >> openclose; openclose=='(' || ')' ;cin >> openclose){
        if (openclose=='('){
            open+=1;
        }
        else if (openclose==')'){
            close+=1;
        }
    }
    if ((open-close)==0){
        cout<<"VALID 0";
    }
    else if ((open-close)!=0){
        int d=open-close;
        cout<<"INVALID"<<d;
    }
}

Grading criterion: Printing Valid and Invalid
Rating A: correctly prints VALID / INVALID and the count based on the condition
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{
    
}

Grading criterion: Printing Valid and Invalid
Rating A: correctly prints VALID / INVALID and the count based on the condition
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 a=0;
        int b=0;
        int n;
        cin>>n;
    while(n>=0)
    {
        char input;
        cin>>input;
        if(input=='(')
        {
            a=a+1;
        }
        if(input==')')
        {
            b=b+1;
        }
    }
    int c=(a-b);
    if (c!=0)
    {cout<<"INVALID "<<c<<endl;}
    else{
        cout<<"VALID "<<c<<endl;
    }
    n--;
}

Grading criterion: Printing Valid and Invalid
Rating A: correctly prints VALID / INVALID and the count based on the condition
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,i=1,opencount=0,closecount=0,s=0,p=0;
    cin>>n;
    for(i=1;i<=n;i++)
    {
        char c;
        cin>>c;
        if (c=='(')
        { 
            opencount=opencount+1;
        }
        if (c==')')
        {
            closecount=closecount+1;
        }
        if ((i==1 && c!='(') || (i==n && c!=')'))
        {
            p=p+1;
        }
    }
    s=opencount-closecount;
    if ( p==0 && s==0)
    {
        cout<<"VALID"<<" "<<s; 
    }
    if (p!=0)
    {
        cout<<"INVALID"<<" "<<s;
    }
    

}

Grading criterion: Printing Valid and Invalid
Rating A: correctly prints VALID / INVALID and the count based on the condition
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 c1=0; int c2=0;bool valid=true;
	cin>>n;
    char c;
    cin>>c;

	for(int i=0;i<n;i++, cin>> c){
	if (c=='(') c1++;
	if (c==')') c2++;
	if (c2>c1 )  valid=false;
	}
	int s=c1-c2;
	if(valid) {if (s==0) cout<<"VALID "<< s; else {valid=false;cout<<"INVALID"<<s;}}
	else cout<<"INVALID "<<s;
}

Grading criterion: Printing Valid and Invalid
Rating A: correctly prints VALID / INVALID and the count based on the condition
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=0;
cin >> n;
int open,close =0;
char c;

if (n==0){
    cout << "VALID 0";
}
int diff = open-close;
bool final = true;
cin >> c;
if (c==')'){
    close++;
    final = false;}
else 
    open++;

for (int i=1;i<n;i++){
    
    cin >> c;
    if (c=='(')
        open++;
    else 
         close++;
    diff = open-close;
    
}



if (open!=close)
    cout << "INVALID "<<diff;
else if (open==close && final==false)
    cout << "INVALID 0";
else 
    cout << "VALID 0";

}

Grading criterion: Printing Valid and Invalid
Rating A: correctly prints VALID / INVALID and the count based on the condition
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 input;
        char '(';
        char ')';
        int i;
        for(i=1; i<=n; i++){
                cin>>input;
        }

Grading criterion: Printing Valid and Invalid
Rating A: correctly prints VALID / INVALID and the count based on the condition
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 {
    unsigned int n;
    cin>>n;
    int open=0,close=0;
    int op=0,cl=0;
    for(int i=0; i<n; i++) {
        char input;
        cin>>input;
        if(input==')') close++;
        else if(input=='(') open++;
    }
    for(int i=0; i<n; i++) {
        char input;
        cin>>input;
        if(input==')') cl++;
        else if(input=='(') op++;
        if(cl>op) break;
        else continue;
    }
    if(op>=cl && open==close) cout<<"VALID 0";
    else cout<<"INVALID"<<" "<<abs(open-close);
}

Grading criterion: Printing Valid and Invalid
Rating A: correctly prints VALID / INVALID and the count based on the condition
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,temp=1,ocount=0,ccount=0;
    cin>>n;
    for(int i=1;i<=n;i++){
        char c;
        cin>>c;
        if(c=='('){ocount++;}
        else{ccount++;}
        if(i<n&&ocount-ccount<0){
            temp=0;
            }
        
    }
    if(temp==0){cout<<"INVALID "<<ocount-ccount;}
    else{
        if(ocount!=ccount){cout<<"INVALID "<<ocount-ccount;}
        else{cout<<"VALID "<<0;}
    }
}

Grading criterion: Printing Valid and Invalid
Rating A: correctly prints VALID / INVALID and the count based on the condition
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A