Alphanumeric Abbreviations of a String
Given a string of characters of length less than 10. We need to print all the alpha-numeric abbreviation of the string. The alpha-numeric abbreviation is in the form of characters mixed with the digits which is equal to the number of skipped characters of a selected substring. So, whenever a substring of characters is skipped, you have to replace it with the digit denoting the number of characters in the substring. There may be any number of skipped substrings of a string. No two substrings should be adjacent to each other. Hence, no two digits are adjacent in the result. For a clearer idea, see the example. Examples:
Input : ANKS Output : ANKS (nothing is replaced) ANK1 (S is replaced) AN1S (K is replaced) AN2 (KS is replaced) A1KS (N is replaced) A1K1 (N and S are replaced) A2S (NK is replaced) A3 (NKS is replaced) 1NKS (A is replaced) 1NK1 (A and S are replaced) 1N1S (A and N is replaced) 1N2 (A and KS are replaced) 2KS (AN is replaced) 2K1 (AN and S is replaced) 3S (ANK is replaced) 4 (ANKS is replaced) Input : ABC Output : ABC AB1 A1C A2 1BC 1B1 2C 3 Note: 11C is not valid because no two digits should be adjacent, 2C is the correct one because AB is a substring, not A and B individually
Source: Google Interview question
The idea is to start with empty string. At every step, we have two choices.
- Consider character as it is.
- Add character to count. If there is no count, then use 1.
You can see how each character can either add up to the result as a character or as a digit. This further gives rise to 2^n abbreviations at the end where n is the length of string.
CPP
// C++ program to print all Alpha-Numeric Abbreviations// of a String#include <bits/stdc++.h>using namespace std;// Recursive function to print the valid combinations// s is string, st is resultant stringvoid printCompRec(const string& s, int index, int max_index, string st){ // if the end of the string is reached if (index == max_index) { cout << st << "\n"; return; } // push the current character to result st.push_back(s[index]); // recur for the next [Using Char] printCompRec(s, index + 1, max_index, st); // remove the character from result st.pop_back(); // set count of digits to 1 int count = 1; // addition the adjacent digits if (!st.empty()) { if (isdigit(st.back())) { // get the digit and increase the count count += (int)(st.back() - '0'); // remove the adjacent digit st.pop_back(); } } // change count to a character char to_print = (char)(count + '0'); // add the character to result st.push_back(to_print); // recur for this again [Using Count] printCompRec(s, index + 1, max_index, st);}// Wrapper functionvoid printComb(std::string s){ // if the string is empty if (!s.length()) return; // Stores result strings one by one string st; printCompRec(s, 0, s.length(), st);}// driver functionint main(){ string str = "GFG"; printComb(str); return 0;} |
Python3
# Python program to print all Alpha-Numeric Abbreviations# of a String# Recursive function to print the valid combinations# s is string, st is resultant stringdef printCompRec(s, index, max_index, st): # if the end of the string is reached if (index == max_index): print(st) return # push the current character to result st += s[index] # recur for the next [Using Char] printCompRec(s, index + 1, max_index, st) # remove the character from result st = st[0:len(st)-1] # set count of digits to 1 count = 1 # addition the adjacent digits if (len(st) > 0): if (ord(st[-1])>=ord('0') and ord(st[-1])<=ord('9')): # get the digit and increase the count count += (ord(st[-1]) - ord('0')) # remove the adjacent digit st = st[0:len(st)-1] # change count to a character to_print = chr(count + ord('0')) # add the character to result st += to_print # recur for this again [Using Count] printCompRec(s, index + 1, max_index, st)# Wrapper functiondef printComb(s): # if the string is empty if (len(s) == 0): return # Stores result strings one by one st = "" printCompRec(s, 0, len(s), st)# driver functionStr = "GFG"printComb(Str)# This code is contributed by shinjanpatra |
Output:
GFG GF1 G1G G2 1FG 1F1 2G 3
Source: CareerCup This article is contributed by Aditya Nihal Kumar Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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