Here we will see an interesting problem related to alphanumeric abbreviation of a given string. String length is less than 10. We will print out all alphanumeric abbreviations.
Alphanumeric abbreviations are formed by a mixture of characters and numbers. The value of this number is the number of characters that were omitted. There may be any number of omitted substrings. No two substrings are adjacent. Let's look at the algorithm for obtaining this concept.
printAbbreviation(s, index, max, str) −
begin
if index is same as max, then
print str
end if
add s[index] at the last of str
printAbbreviation(s, index + 1, max, str)
delete last character from str
count := 1
if str is not empty, then
if the last character of str is a digit, then
add last digit with the count value
delete last character from str
end if
end if
add count after the str
printAbbreveation(s, index + 1, max, str)
end#include <iostream>
using namespace std;
void printAbbreviation(const string& s, int index, int max_index, string str) {
if (index == max_index) { //if string has ended
cout << str << endl;
return;
}
str.push_back(s[index]); // push the current character to result
printAbbreviation(s, index + 1, max_index, str); //print from next index
str.pop_back(); //remove last character
int count = 1;
if (!str.empty()) {
if (isdigit(str.back())) { //if the last one is digit, then
count += (int)(str.back() - '0'); //count the integer value of that digit
str.pop_back(); //remove last character
}
}
char to_char = (char)(count + '0'); //make count to character
str.push_back(to_char);
printAbbreviation(s, index + 1, max_index, str); //do for next index
}
void printCombination(string str) {
if (!str.length()) //if the string is empty
return;
string str_res;
printAbbreviation(str, 0, str.length(), str_res);
}
int main() {
string str = "HELLO";
printCombination(str);
}HELLO HELL1 HEL1O HEL2 HE1LO HE1L1 HE2O HE3 H1LLO H1LL1 H1L1O H1L2 H2LO H2L1 H3O H4 1ELLO 1ELL1 1EL1O 1EL2 1E1LO 1E1L1 1E2O 1E3 2LLO 2LL1 2L1O 2L2 3LO 3L1 4O 5
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