Decode String

Probelm :
You are given an encoded string that needs to be decoded following a specific set of rules. The encoded pattern is formatted as k[encoded_string], where k signifies the number of times the encoded_string should be repeated. This means the value within the square brackets should be repeated k times to decode it. The encoded string is guaranteed to be a valid format, with no extra spaces, and properly balanced square brackets. Additionally, any numbers in the string refer directly to the repetition count.

Example:
Consider the input string '3[a2[c]]'.

Start with the innermost brackets 2[c]. Here c is repeated twice to become cc.
Replace 2[c] with cc, resulting in the string '3[acc]'.
Decode the remaining pattern, repeating acc three times to yield accaccacc.
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class Solution {
public:
    string decodeString(string s) {
       stack<string> st;

       for(auto ch : s){
        if(ch == ']'){
            string toRepeat = "";
            while(!st.empty() && st.top() != "["){
                string top = st.top();
                toRepeat += top;
                st.pop();
            }
            //to pop ']' so that we can get the numeric digits
            st.pop();
            string numericDigit ="";
            while(!st.empty() && isdigit(st.top()[0])){
                numericDigit +=st.top();
                st.pop();
            }

            //reverse to get the actual number
            reverse(numericDigit.begin(),numericDigit.end());
            //convert string to int to get the number
            int n = stoi(numericDigit);

            string currDecode = "";
            while(n -- ){
               currDecode += toRepeat;
            }
            st.push(currDecode);
        }
        else{
            string temp(1,ch);
            st.push(temp);
        }

       }

       //now all the final decoded string are in the stack
       string ans ="";
       while(!st.empty()){
        ans += st.top();
        st.pop();
       }

       //as its a stack we need to reverse the final ans toget the actual ans

       reverse(ans.begin(),ans.end());
       return ans;
    }
};