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# 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

The idea is to start with an empty string. At every step, we have two choices.

1. Consider character as it is.
2. 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.

Implementation:

## CPP

 // C++ program to print all Alpha-Numeric Abbreviations // of a String #include using namespace std;   // Recursive function to print the valid combinations // s is string, st is resultant string void 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 function void 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 function int main() {     string str = "GFG";     printComb(str);     return 0; }

## Java

 import java.io.*;     public class GFG {                 // Recursive function to print the valid combinations     // s is string, st is resultant string     static void printCompRec(String s, int index, int max_index, String st)     {         // if the end of the string is reached         if (index == max_index) {             System.out.println(st);             return;         }           // push the current character to result         st = st + s.charAt(index);           // recur for the next [Using Char]         printCompRec(s, index + 1, max_index, st);           // remove the character from result         st = st.substring(0, st.length() - 1);           // set count of digits to 1         int count = 1;           // addition the adjacent digits         if (st.length() > 0) {               if (st.charAt(st.length()-1) >= '0' && st.charAt(st.length() -1) <= '9') {                   // get the digit and increase the count                 count = count + (st.charAt(st.length()-1) - '0');                   // remove the adjacent digit                 st = st.substring(0, st.length() - 1);             }         }           // change count to a character         char to_print = (char)(count + '0');           // add the character to result         st = st + to_print;           // recur for this again [Using Count]         printCompRec(s, index + 1, max_index, st);     }           // Wrapper function     static void printComb(String s)     {         // if the string is empty         if (s.length() == 0)             return;           // Stores result strings one by one         String st = "";           printCompRec(s, 0, s.length(), st);     }           public static void main(String[] args) {           String str = "GFG";         printComb(str);     } }   // The code is contributed by Nidhi goel.

## 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 string def 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 function def 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 function Str = "GFG" printComb(Str)   # This code is contributed by shinjanpatra

## C#

 using System;   // C# program to print all Alpha-Numeric Abbreviations // of a String class GFG {     // Recursive function to print the valid combinations   // s is string, st is resultant string   public static void printCompRec(string s, int index, int max_index, string st)   {     // if the end of the string is reached     if (index == max_index) {       Console.WriteLine(st);       return;     }       // push the current character to result     st = st + s[index];       // recur for the next [Using Char]     printCompRec(s, index + 1, max_index, st);       // remove the character from result     st = st.Substring(0,st.Length-1);       // set count of digits to 1     int count = 1;       // addition the adjacent digits     if (st.Length > 0) {         if ((int)st[st.Length-1] >=(int)'0' && (int)st[st.Length - 1]<= (int)'9') {           // get the digit and increase the count         count += ((int)st[st.Length-1] - (int)'0');           // remove the adjacent digit         st = st.Substring(0,st.Length-1);       }     }       // change count to a character     char to_print = (char)(count + '0');       // add the character to result     st = st + to_print;       // recur for this again [Using Count]     printCompRec(s, index + 1, max_index, st);   }       // Wrapper function   public static void printComb(string s)   {     // if the string is empty     if (s.Length == 0)       return;       // Stores result strings one by one     string st = "";       printCompRec(s, 0, s.Length, st);   }       static void Main() {     string str = "GFG";     printComb(str);   } }   // The code is contributed by Nidhi goel.

## Javascript



Output

GFG
GF1
G1G
G2
1FG
1F1
2G
3

Time Complexity: O(2^N), Where N is the length of the input string.
Auxiliary Space: O(2^N)

### We can generate all these Abbreviations Iteratively too

Algorithm : 1. Let string of length n = 3 , str = “ABC” binary value between 0 to 7 will helps in deciding which character to be used or which character will be replaced If for let say n = 6 , binary = 110 consider each binary bit position as index bit = 1 means it will be replaced and 0 means we are not changing that index character and 0 means it remain as it is

Implementation:

## C++

 #include using namespace std;     void printans(string str) {     string ans = "";     int counter = 0;     for (auto ch: str) {         if (ch == '-') {             counter++;         }         else {             if (counter > 0)                 ans = ans + to_string(counter);             counter = 0;             ans = ans + ch;         }     }     if (counter > 0)         ans = ans + to_string(counter);     cout << ans << endl; }   int main() {       string str = "ANKS";       int n = str.size();     int limit = 1 << n;     for (int i = 0; i < limit; i++) {         int counter = i, idx = 0;         string abb = "";         for (int b = 0; b < n; b++) {             int bit = counter % 2;             counter /= 2;             if (bit == 1) {                 abb = abb + "-";             }             else {                 abb = abb + str[idx];             }             idx++;         }         printans(abb);     }           return 0; }   // The code is contributed by Gautam goel.

## Java

 import java.io.*; import java.util.*;   public class Main {       private static void printans(String str)     {         String ans = "";         int counter = 0;         for (char ch : str.toCharArray()) {             if (ch == '-') {                 counter++;             }             else {                 if (counter > 0)                     ans = ans + counter;                 counter = 0;                 ans = ans + ch;             }         }         if (counter > 0)             ans = ans + counter;         System.out.println(ans);     }       public static void main(String[] args)     {           String str = "ANKS";           int n = str.length();         int limit = 1 << n;         for (int i = 0; i < limit; i++) {             int counter = i, idx = 0;             String abb = "";             for (int b = 0; b < n; b++) {                 int bit = counter % 2;                 counter /= 2;                 if (bit == 1) {                     abb = abb + "-";                 }                 else {                     abb = abb + str.charAt(idx);                 }                 idx++;             }             printans(abb);         }     } }

## Python3

 # Python code for above approach def printans(Str):     ans=""     counter=0     for ch in Str:         if(ch=='-'):             counter+=1         else:             if(counter>0):                 ans+=str(counter)             counter=0             ans+=ch           if(counter>0):         ans+=str(counter)           print(ans)       # Str = "ANKS"   n=len(Str) limit=1<

## C#

 // C# code for above approach using System;   public class GFG{       private static void printans(String str)     {         string ans = "";         int counter = 0;         foreach (char ch in str) {             if (ch == '-') {                 counter++;             }             else {                 if (counter > 0)                     ans = ans + counter;                 counter = 0;                 ans = ans + ch;             }         }         if (counter > 0)             ans = ans + counter;         Console.WriteLine(ans);     }       public static void Main()     {           string str = "ANKS";           int n = str.Length;         int limit = 1 << n;         for (int i = 0; i < limit; i++) {             int counter = i, idx = 0;             string abb = "";             for (int b = 0; b < n; b++) {                 int bit = counter % 2;                 counter /= 2;                 if (bit == 1) {                     abb = abb + "-";                 }                 else {                     abb = abb + str[idx];                 }                 idx++;             }             printans(abb);         }     } }   // This code is contributed by Aman Kumar

## Javascript

 // JavaScript program for the above approach function printans(str){     let ans = "";     let counter = 0;     for(let i = 0; i 0) ans = ans + counter.toString();             counter = 0;             ans = ans + ch;         }     }     if(counter > 0) ans = ans + counter.toString();     console.log(ans); }   // driver program for the above function let str = "ANKS"; let n = str.length; let limit = 1<

Output

ANKS
1NKS
A1KS
2KS
AN1S
1N1S
A2S
3S
ANK1
1NK1
A1K1
2K1
AN2
1N2
A3
4

Time Complexity: O(2^N), Where N is the length of the input string.
Auxiliary Space: O(N), for storing strings in a temporary string.

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.

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