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.
// 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 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 one by one string st; printCompRec(s, 0, s.length(), st); } // driver function int main() { string str = "GFG"; printComb(str); return 0; } |
Output:
GFG GF1 G1G G2 1FG 1F1 2G 3
Source: CareerCup
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