Java Interview Question: How To Count Occurrences Of Each Character In String In Java

All my interest in Java is because of you. I can't thank you enough for all that you have done for us!! 👏

The people who getting error in isBlank method they can use in that if statement like below
if(c!=' '){



}

solution for counting space
int count=0;
for(char ch:name.toCharArray()) {
if(ch==' ') {
count++;
}
}
System.out.println("******************");
System.out.println(count);

Solution for only space we need to take count of string and replaceall method
int n=name.length()-name.replaceAll(" ", "").length();
System.out.println("spaceCount");
System.out.println(n);

Sir, how can i get output as te2s3h if input is teesssh with stream ??

sir ap hindi main bnao yr video

YOU'RE THE MAN!!, THANK YOU!!!!!!!!!!!! but why u had to test ur code for 10 min lol

To count only spaces, we will remove NOT from condition like below
if(String.valueOf(c).isBlank())

veryyy very nice explaination

simple remove ! in (String.valueOf(C).isBlank()){} condtion

Beautiful explanation

Print occurences of each word and character in a string : it was asked in the interview

Thank you Naveen!! God bless you 🙏

Hi Naveen, I m getting error while using the isBlank() method. Error showing that method isBlank() is undefined for type string.

Yes sir, by just removing the (!) In the
if(!string.valueof(c).isBlank()) we got only spaces

String S1="hello"
String S2="12345"
Output=h1e2l3l4o5. Can you plz write program for above output

sir. thanks for the video .can we do this in java 8

Better, u can do videos for Less time

1 2 three 3 4 three 5 6 three 7 8 three.......20.anyone can solve this please aisa output aana chahiye

Your approach is really so easy to understand and you solve the program in most efficient way too
Thank you so much...

Hi Naveen,
Just a question..

For thr first test case i.e input "test" the output Hashmap values must be {(t,1)(e,1)(s,1)(t,2)} .. right??
Because we are just adding a new key value pair to the hashmap but we are not updating the existing key-value pairs as per this code...
For the 4th input character : 't' it will check if 't' is already present in the map. Since it is already there, it will fetch it's value which is 1 and then increments it by 1..so it will become 2 and will update a new key-value pair as( t,2) to the existing hashmap...so the final hashmap output should be {(t,1)(e,1)(s,1)(t,2)} ..
Please explain how it is giving {(e,1)(s,1)(t,2)} as output..

Thanks in advance.