length of last word Algorithm
Additionally, comparisons are complicated because place names may be considered words, technical terms may be randomly long, and the addition of suffixes and prefixes may extend the length of words to make grammatically correct but unused or novel words. The identity of the longest word in English depends upon the definition of what constitutes a word in the English language, as well as how length should be compared.
/*
* Given a string consisting of lower/upper case alphabets and empty space character ' '.
* Return the length of last word in the string.
*
* A word is character sequence containing non-space characters only.
*
* Return 0, if no last word exist.
*
* Approach:
*
* Start from end of the string, and trim the whitespace.
* When we find the first non-space character, find the next space or till we reach beginning of string.
*
*/
#include <iostream>
int get_length_of_last_word(const std::string& str)
{
if (str.length() == 0) {
return 0;
}
int n = str.length();
int i = n - 1;
int j = 0;
// Trim the whitespace in the end
while(i >= 0 && str[i] == ' ') i--;
// Check if we already reached beginning of the string.
if (i < 0) {
return 0;
}
// Now start from the first non-space character
for (j = i; j >= 0; --j) {
if (str[j] == ' ') break;
}
return i - j;
}
int main()
{
std::string str1 = "Hello";
std::string str2 = "Hello ";
std::string str3 = "Hello World!";
std::cout << "String without any space of length " << str1.length() << ": " << str1 << std::endl;
std::cout << "Length of last word :" << get_length_of_last_word(str1) << std::endl;
std::cout << "String with lot of space in the end, with length of string " << str2.length() << ": " << str2 << std::endl;
std::cout << "Length of last word :" << get_length_of_last_word(str2) << std::endl;
std::cout << "String with lot of space in the middle, with length of string " << str3.length() << ": " << str3 << std::endl;
std::cout << "Length of last word :" << get_length_of_last_word(str3) << std::endl;
return 0;
}
