Problem:
Given a string, count the number of words ending in 'y' or 'z' -- so the 'y' in "heavy" and the 'z' in "fez" count, but not the 'y' in "yellow" (not case sensitive). We'll say that a y or z is at the end of a word if there is not an alphabetic letter immediately following it. (Note: Character.isLetter(char) tests if a char is an alphabetic letter.)
countYZ("fez day") → 2
countYZ("day fez") → 2
countYZ("day fyyyz") → 2
Solution:
public int countYZ(String str) {
int len = str.length();
int count = 0;
str = str.toLowerCase();
for (int i = 0; i < len; i++) {
if (str.charAt(i) == 'y' || str.charAt(i) == 'z') {
if (i < len-1 && !Character.isLetter(str.charAt(i+1)))
count++;
else if (i == len-1)
count++;
}
}
return count;
}
public int countYZ(String str) {
ReplyDeletestr = str.toLowerCase();
int c=0;
for(int i=0; i<str.length()-1; i++){
if((str.charAt(i)=='y' || str.charAt(i)=='z') && !Character.isLetter(str.charAt(i+1)))
c++;
}
if(str.charAt(str.length()-1)=='y' || str.charAt(str.length()-1)=='z')
c++;
return c;
}
My solution is very similar, except that I first check if the string ends with 'y' or 'z'. This simplifies the for loop a bit:
ReplyDeletepublic int countYZ(String str) {
int count = 0;
str = str.toLowerCase();
if(str.endsWith("y") || str.endsWith("z")){
count++;
}
for(int i = 0; i < str.length() - 1; i++){
if( (str.charAt(i) == 'y' || str.charAt(i) == 'z') &&
!(Character.isLetter(str.charAt(i+1))) ){
count++;
}
}
return count;
}
public int countYZ(String str) {
ReplyDeletestr = str.toLowerCase();
String[] words = str.split("[^a-z]");
int result = 0;
for (int i = 0; i < words.length; i++)
if (words[i].endsWith("y") || words[i].endsWith("z"))
result++;
return result;
}
split("[^a-z]"); can you explain, why you use ^
Deletepublic int countYZ(String str) {
ReplyDeleteint count = 0;
for (int i = str.length() - 1; i >= 0; i--)
{
if ((Character.toLowerCase(str.charAt(i)) == 'y' || Character.toLowerCase(str.charAt(i)) == 'z') && (i == str.length() - 1 || !Character.isLetter(str.charAt(i + 1))))
{
count++;
}
}
return count;
}
you gonna lol on my dirty logic
ReplyDeletepublic int countYZ(String str) {
str=str.toLowerCase()+" ";
int count=0;
for(int i=0;i<str.length()-1;i++){
if(str.charAt(i)=='y' || str.charAt(i)=='z'){
if(!Character.isLetter(str.charAt(i+1))){
count++;
}
}
}
return count;
}
public int countYZ(String str) {
ReplyDeleteint c=0;
int t=str.length()-1;
str = str.toLowerCase();
for(int i=0;i<str.length();i++){
if(str.charAt(i)=='y'||str.charAt(i)=='z')
{
if(i < t && !Character.isLetter(str.charAt(i+1)))
c++;
else if (i == t)
c++;
}}return c;
}
im doin all of this in python cuz java has more tasks, here u go if u are wonder :D
ReplyDeletedef countYZ(s):
s = s.lower()
c = 0
if s[len(s)-1] == "y" or s[len(s)-1] == "z":
c += 1
for i in range(len(s)):
if not s[i].isalpha() and s[i-1] == "y":
c += 1
if not s[i].isalpha() and s[i-1] == "z":
c += 1
return c
The main solution at the top works better, but here is another logic for everyone.
ReplyDeletepublic int countYZ(String str) {
int countYZ = 0;
str = str.toLowerCase();
for (int i = 0; i < str.length(); ++i) {
boolean yz = (str.charAt(i) == 'y' || str.charAt(i) == 'z');
boolean endWord = (i + 1 >= str.length() || !(Character.isLetter(str.charAt(i+1))));
if (yz && endWord) countYZ++;
}
return countYZ;
}
With Regex
ReplyDeletepublic int countYZ(String str) {
return str.split("((?i)z[^a-zA-Z]|z$)|((?i)y[^a-zA-Z]|y$)", - 1).length - 1;
}