If user enters a string, i want to make a function to check if a repetition of a certain number of zeroes follows an equal number of ones. Example: 001101 (correct), 01(correct),001010 (incorrect). I've tried to store the string in 2 separate strings and compare the size but my 2nd while loop isn't stopping.

void check(string num) {
    string st0 = "", st1 = "";
    int n = num.length();
    int k = 0;
    
    while (k < n) {
        int i = 0;

        while (num[i] == num[i 1]) {    
            st0.push_back(num.back());
            num.pop_back();
            k  ;
            i  ;
        }
    
        st0.push_back(num.back());
        num.pop_back();
        k  ;
    
        int j = 0;
        while (num[j] == num[j 1]) {
            st1.push_back(num.back());
            num.pop_back();
            k  ;
            j  ;
        }

        st1.push_back(num.back());
        num.pop_back();
        k  ;

        if (st0.size() != st1.size()) {
            cout << "incorrect \n";
        }

        st0.clear();
        st1.clear(); 
    }
}

CodePudding user response：

Here you are testing the elements from left to right and you're adding to st0 the last elements from num string. For example:
"111010": after the first while num = "111" and st0 = "010"

 while (num[i] == num[i 1]) {    
            st0.push_back(num.back());
            num.pop_back();
            k  ;
            i  ;
        }

And to fix you should add the first element tested not the last.

while( num[i] == num[i 1] && i 1 < n ){ 
                st0.push_back(num[i]);
                i  ;
            }
            st0.push_back(num[i]); 
            i  ;

You did the same mistake here

while (num[j] == num[j 1]) {
            st1.push_back(num.back());
            num.pop_back();
            k  ;
            j  ;
        }

Fix:

while( num[i] == num[i 1] && i 1 < n){
             st1.push_back(num[i]); // same here 
             i  ;
        }
        st1.push_back(num[i]);
        i  ;

I used the same i here so that we continue looping the string from where we left off in the first while.
But there is a more optimized solution if you're interested:

void check(string num)
{
    int cnt0 = 0 , cnt1 = 0;
    int n = num.length();
    int i=0;
    bool test = true ; 

    while (i<n){
       /* we can use two variables and increment them instead of strings and compare them cauz we are not using the elements we just need the size of them */
            while( num[i] == num[i 1] && i 1 < n ){ 
                cnt0   ;
                i  ;
            }
            cnt0  ;
            i  ;
            while( num[i] == num[i 1] && i 1 < n) {
              cnt1  ;  
              i  ;
            }
            cnt1  ;
            i    ;
            if ( cnt1 != cnt0 ){
                 test = false ; 
                 break; 
            }
            cnt1 = 0;
            cnt0 = 0;
    }
    if( test ) { 
         cout << "correct \n" ;
    }
    else {
        cout << "incorrect \n" ; 
    }
}

In the above algorithm, as you can see, you only need the length of the string st0 and st1 so you could just use 2 int variables and test the difference between the two, this way is more memory efficient as well as slightly faster run times.