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Alphanumeric Abbreviations of a String in C Program?
Here we will see one interesting problem related to alphanumeric abbreviation of a given string. The string length is less than 10. We will print all alphanumeric abbreviations.
The alphanumeric abbreviation is in the form of characters mixed with the digits. The value of that digit is number of characters that are missed. There may be any number of skipped substrings. No two substrings are adjacent to each other. Let us see the algorithm to get the idea.
Algorithm
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
Example
#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);
}Output
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
Updated on: 20-Aug-2019
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