message: solution added, with logics
diff: +/* The program runs a main loop to iterate over a string character by character
  + The Complexity of the code is O(n).
  +
  + The string is iterated character by character, and in each iteration the
  + i+2th character is checked if it is #. If it is hash then the string considered is
  + XX# and we convert XX to its equivalent integer. Otherwise the character is changed to its ascii value
  + and 48 is subtracted from it.
  +
  + Following this we add 96 in order to reach the ascii value of the corresponding ascii of the characters.
  + ASCII values:
  + a -> 97
  + A -> 65
  + 1 -> 49
  +
  + The ascii values are appended to a vector<int> and later converted to the equivalent string
  +*/
  +
  +class Solution {
  +public:
  + string freqAlphabets(string s) {
  + vector<int> answer;
  + int i, string_length = s.length();
  + for(i=0; i<string_length-1; i++){
  + if(s[i+2] == '#'){
  + answer.push_back((((int)(s[i])-48)*10 + (int)(s[i+1])-48)+96);
  + i+=2;
  + }
  + else
  + answer.push_back(((int)(s[i])-48)+96);
  + }
  + if(s[string_length-1] !='#') answer.push_back(((int)(s[i])-48)+96);
  + char mapped_string[answer.size()+1];
  + for(int j=0; j<answer.size(); j++){
  + mapped_string[j] = (char)(answer[j]);
  + }
  + mapped_string[answer.size()] = '\0';
  + return mapped_string;
  + }
  +};
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