Leetcode License Key Formatting problem solution

In this Leetcode License Key Formatting problem solution You are given a license key represented as a string s that consists of only alphanumeric characters and dashes. The string is separated into n + 1 groups by n dashes. You are also given an integer k.

We want to reformat the string s such that each group contains exactly k characters, except for the first group, which could be shorter than k but still must contain at least one character. Furthermore, there must be a dash inserted between two groups, and you should convert all lowercase letters to uppercase.

Return the reformatted license key.

Problem solution in Python.

class Solution(object):
    def licenseKeyFormatting(self, S, K):
        s = S.replace('-', '').upper()
        ngroups, rem = len(s) / K , len(s) % K
        S = s[0:rem]
        for i in range(ngroups):
            if i > 0 or rem > 0: S += '-'
            S += s[rem+i*K:rem+(i+1)*K]
        return S

Problem solution in Java.

public String licenseKeyFormatting(String S, int K) {
        StringBuilder sb = new StringBuilder(); 
        S = S.replace("-", ""); 
        S = S.toUpperCase();
        if (S.length() == 0) {
            return S; 
        }
        int i = S.length() - 1; 
        while (i >= 0) {
            int count = 0; 
            while (count < K && i >= 0) {
                count++; 
                sb.append(S.charAt(i--)); 
            }
            sb.append("-"); 
        }
        
        sb.deleteCharAt(sb.length() - 1); 
        return sb.reverse().toString(); 
    }

Problem solution in C++.

string licenseKeyFormatting(string S, int K) {
        if (S.empty()) return S;
 
        std::string key;
        key.reserve(2*S.length());
        int len = 0;
        for_each(S.rbegin(), S.rend(), [&](char c)
        {
            if (c != '-')
            {
                if(len == K)
                {
                    key += "-";
                    len = 0;
                }
                ++len;
                key += std::toupper(c);
            }
        });
        std::reverse(key.begin(), key.end());
        return key;
        
    }