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

In this video, I have explained one famous Java Interview Question: How To Count Occurrences Of Each Character In String In Java
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[@sanghamitrabhowmicklive3013]

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

[@surabhibn541]

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

[@lavanya.ds0657]

Thank you souch for sharing this Naveen, this is mandatory question asked in many interviews for experienced candidates

[@909praveen]

Thanks Naveen for the video . For the assessment we just have to remove not symbol from if(String.valueOf(c).is blank()).

[@somrajyadav2446]

Thank you Naveen!! God bless you 🙏

[@shashikumarkondoju7808]

Thankyou sir the way you explained awesome..it's makes every one to easy understand...
For only spaces count we use instead of not if condition
if(String.valueOf(ch).isBank())

[@khajazakiuddin931]

Thank you so much for the amazing videos on Java concepts Naveen. it really helps in improving my coding skills. Could you also please make a video on how to print all the permutations of a given string "abc" (like acb, bca, cab, cba..etc.,). (It was asked in the interview with ADP company). There are many logics present on the internet but not fully understandable, just wanted to get it from you which obviously would be easy to understand and implement.

[@vigneshlakshmipathi2689]

Thank you so much naveen😍

[@NagarajuN-ev3pl]

Hi NAVEEN,
Good day!
The same program I was facing in the technical round and I wrote and explained to the interviewer. I cracked and cleared the first round as well. Thank you, please post more mock interviews. Thanks a lot..

[@varshamishra8743]

Thank you Sir.. Nice video 👍😊

[@vijaykanthperikala]

Hi Naveen, I'm getting below error when I try to use "is Blank()" method.
"The method isBlank() is undefined for the type String"
Could you please help me ?
I'm using jdk 1.8.0-301 version

[@sayyedfiroz7556]

Very nice video Naveen

[@13success66]

Beautiful explanation

[@arunkumar-ke3km]

veryuseful video,

[@aayushpatel3360]

Best approach

[@ShashiBhushanSrivastava001]

Thanks Naveen.

[@ashuiet]

Thanks!

[@anantpandey309]

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

[@VivekKumar-bu6ho]

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

[@parthjadhav2501]

veryyy very nice explaination

[@anilds2631]

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);

[@thatsdeep7058]

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

[@antarabanerjee7314]

Thanks for the explanation Naveen. It's because of you, yesterday I cracked one more interview!

[@naveenautomationlabs]

Many congratulations 👍

[@antarabanerjee7314]

@@naveenautomationlabs thank you so much!

[@sruthibhagavatula2450]

Why did you give only hash map? And how to choose which map to use in which scenario?

[@ParamjitKaur-jo2sz]

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

[@devanandthirunavukkarasu5148]

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

[@lathaa890]

Make a video for String Builder and String Buffer bro.

[@ashuiet]

most of the interview this question asked ...if not this then ...some what similar to this problem.

[@haribabuvelakavaram5935]

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.

[@SurajSharma-fw5fg]

Because he is using HashMap and HashMap can not contain duplicate "key". Hence, only one 't' will be there in HashMap.

[@anilds2631]

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

[@Wandereatstravel]

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

[@aparna5088]

This program is case sensitive
the uppercase and lowercase occurences of same letters are counted seperately.

[@MrAshu58]

I am well versed in Python, Should I learn Java as well?

[@naveenautomationlabs]

Yes if you trying for Selenium. Selenium with java is the most demanding as compared to other language bindings.

[@ananthvraj183]

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

[@mohan4459]

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

[@lalitjadaunreaction014]

bhai simple hai 1 loop creta karlo then bhs ek character fist string sai or 2nd character from 2 string using charAt[i]. simple sa trika to ye hai . or bhi efficienct way

[@district_gamer]

sir ap hindi main bnao yr video

[@saikumar-io7un]

Better, u can do videos for Less time

[@STUDIES-wv4mw]

public static void spaceCount(String str) {
int count = 0;
char[] c = str.toCharArray();
for (char ce : c) {
if (ce == ' ') {
count++;
}
}
System.out.println("Space Count " + count);
}

[@mohankumarrewatkar2824]

Hello Naveen thanks for this video..
One interview que...if u can try for us...
String str[]={"Suresh:20","Ramesh:10","Anand:05"}
We want to sort the str array on the basis of age....

[@mertm2245]

String str[]={"Suresh:20","Ramesh:10","Anand:05"};
List list= new ArrayList();
for(String s:str){
list.add(Integer.parseInt(s.substring(str[0].indexOf(":")+1)));
}
Collections.sort(list);
String str2[]=new String[str.length];
for(int i=0;i

[@codingsskills.explore]

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

[@utkarshpatil2674]

Answer to the assignment, please do correct otherwise
import java.util.HashMap;
import java.util.Map;
public class charCount {

public static void getCharCount(String name) {

Map charMap=new HashMap();

char strArray[]=name.toCharArray();

for(char c: strArray) {
if(String.valueOf(c).contains(" "))
{
if(charMap.containsKey(c)) {
charMap.put(c , charMap.get(c)+1 );
}
else {
charMap.put(c, 1);
}
}
}
System.out.println(name + " : " + charMap);
}
public static void main(String[] args) {
// TODO Auto-generated method stub

getCharCount("sel sel sel");


}
}

[@ravi54217]

public class CharCount {
public static void main(String[] args) {
System.out.println(countBy(" This is a string ", Character::isSpaceChar));
}
private static long countBy(final String string, final IntPredicate condition) {
return string.chars().filter(condition).count();
}
}

[@programminghubbharat7535]

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