Important Interview Question: How to Print count of duplicate characters from String? ||

[@sridharlanka2721]

Truly an "Approach" for Problem solving , and i just learned it!

[@AshutoshMishra-qc7nl]

I got this question in TechnoManagerial interview today only. But I didnt solved in this way. I used your previous tutorial login, did a split of string and kept it in a list, ran 2 for loops for i and i+1 checking and printing any repetitive keywords. Will keep this also in mind

[@santoshkumarvlogs3753]

can you put that solution here @ashutoshMishra

[@bhuvanabaskaran8063]

Crystal and clear explanation... Thank you so much Naveen..keep up the great work.. wishing you all success...you will be blessed forever...Thanks a tonneeee

[@Tidda420]

We can use ctrl + 1 + Enter after ; to get automatic assignment variable but this is applicable only if using eclipse otherwise we have to write it on our own.

[@vishnuchandar7073]

Naveen, as you are the Master of Masters. Thanks for helping me alot in learning. If i want to know what else should i study in Automation , i dont need Google, i just need to visit your youtube page. Thanks. An alternative way to solve the same. Set set = new HashSet(); for(char ch : str.toCharArray()) if(StringUtils.countMatches(str, ch) > 2 && !set.contains(ch)) { System.out.println(ch); set.add(ch); } System.out.print(set);

[@deepakchandrashekar1023]

The question is "Print duplicate characters from String" but the solution also contains the count of duplicate characters in a given string 👍

[@tapaskhandai]

In that case you can only print keys not values

[@WMohanty]

what an explanation. God bless you Naveen...

[@abhishekshandilya4052]

I have paused the video to thankyou, loved the way you have explained.

[@paladarsh39]

Thanks solving this, was asked the same question with a extra condition which was that I have to print the map in the descending order of the repetitive elements.

[@shirishan4525]

for how many year exp you were asked this question??

[@VinayakTungal]

Good approach! 👍 Another case that might be worth considering, is the way white-spaces are treated in this construct. Spaces are neither empty, nor null. So if it's a string full of spaces, what would it be treated as.

[@souvikchatterjee7400]

This is the question I have come across in interview multiple times. Thing is in some cases I write the main logic but forget to handle the edge cases. Still they gave me the interview clearance 😐

[@naveenautomationlabs]

edge cases will always give you bonus points.

[@vikassinghal9413]

This worked as well public static void printviaHashMap(String inputString) { inputString=inputString.toLowerCase(); Map dup = new HashMap(); for(String ch:inputString.split("")) { Integer count = dup.get(ch); dup.put(ch, count==null?1:count+1); }System.out.println(dup); }

[@ayushgupta8239]

Why not just use syso(map) or arrays.aslist or collection? Also I think keyset would be enough to print the keys one by one and then using key to print the value.

[@maywellbeaghost6364]

Good method of solve this particular problem. But, I feel how fast this will go as an obsolete process

[@maneeshbuddha8253]

Hi Naveen,Please do the video on these 2 programs.Java program to print alternate odd numbers which are divisible by 3 and 7 in between 100 to 1000.Java program to print print fibonacci series in reverse order starting from 1000 to 100.

[@phanikumarvishnubhotla1804]

For first program mate public static void main(String[] args) { int count=0,count1=0; for(int i=100;i

[@MrAsgar123]

Thank you so much for your easy explaination.

[@anushap2139]

I really like your voice 😊

[@sonumishra6747]

I think 2 for loop and a string and int max count would do the job. No need of maintaining map set iterating again on it.

[@naveenautomationlabs]

Two for loops will give you O(n^2). But ya would like to see your solution how would you get the max count?

[@ShinAkuma]

@@naveenautomationlabs Just take a integer array of 26 size and update the (char-65)th index on every iteration. You can get the duplicates in O(n+26) complexity

[@vengateshm2122]

Thank you. Time complexity? Space complexity? It would be helpful if you could give a hint on the above for different interview questions.

[@fazex4185]

Time complexity I suppose n + k where k is the number of unique characters.

[@amanuellebassi7294]

Great technique

[@sweetthirty2]

Sir how to print "I love java" into "java love I" plzzz make a video on this frequently getting this problem

[@tapaskhandai]

Hi Naveen, please try to start a series of c# with selenium

[@tapaskhandai]

Hi Naveen, in this case how we can print the 1st repeatative character?

[@prasanthraviv]

Thanks Naveen

[@harinig2874]

Naveen , question is like how come hashmap storing duplicate key ? like a,1 ,a,2,a,3

[@naveenautomationlabs]

Watch this video kzbin.info/www/bejne/iYnJpHV6idSXprs

[@sssaamm29988]

Nice video.should we not use if else conditon? I see that you have used multiple if's.

[@naveenautomationlabs]

It doesn't matter here. We are returning void return from each if condition.

[@jeyhunaliyev1765]

Second here 👍😄✊

[@shivashanker1315]

Hi Naveen, can you help me with the below one. Remove the duplicates from the string, Example: S=”cbbadaabaaebccd”, Condtions: on the 1st iteration, Highest repetitive letter should be removed like “cbbdbebccd”, 2nd iteration, Highest repetitive letter should be removed like “cbbdbebccd” and Final Output should be “e”.

[@RohanKumar-bo6fi]

Hello sir, one doubt I have So when we want to print those characters which are repeated 3 times only then we just need to pass in the if block as: If(entry.getValue==3) Sop(); Is it fine sir?

[@suprabathj7843]

Nice explanation

[@Jyotigupta-vs4mz]

Big thanks

[@ShinAkuma]

Here's and alternate way to do it if you're not allowed to use HashMap. String s = "ABCDEADEF"; int chr[] = new int[26]; for(int i=0; i

[@pandudamera7211]

Please make video alternative numbers and repeat numberss brother

[@pandudamera7211]

Thanks you brother

[@jptechnoservice9013]

Superb bro

[@krishanuchakraborty3500]

Hi Naveen, can't we use HashSet here as well to solve it, instead of HashMap? Any particular reason you advised to use HashMap? Thanks!

[@Erlisch1337]

How would you keep track of the count with a Set? But then again the question is to print duplicate characters, not count them. So I guess a Set should work here too.

[@saikrishna-wi2kg]

hi,how to ignore spaces while finding duplicate count

[@arunprasanth8881]

But in zoho they ask differently...like for example...INPUT: aabbaccddbe OUTPUT: abacdbe How to bring like this?

[@ghyanmitrajiblapnor444]

Why you used set you can directly print the map using for each loop.

[@naveenautomationlabs]

How will you print using simple loop? Map doesn't maintain the order. And you need getvalue()>1.

[@sonyjoseph1498]

​@@naveenautomationlabs for(Map. Entryentry:charMap. EntrySet()){ System.out. println(entry. getKey()+":"+entry.getValue());} can we use this method to print the resul without using set????

[@atifpall2007]

We did not check for capital N versus small n using the string Naveen

[@naveenautomationlabs]

You can check, this will surely work.

[@Amitpatil-g8g]

i think this program is very big naveen sir. is it available another way short.

[@explorer0709]

I want to display as duplicate and non duplicate values what to do in that case

[@priyankshah4027]

U can use streams and partitioning

[@ILuvBilli]

Hi, this is old approach, now a days interviewer wants the solution through stream only

[@abhishekanand3036]

Hi Naveen, I want same result. String st[] = {"Naveen","Java","Tom","A","",null}; Help me here.... Also share without using collection

[@rdnkmr]

iterate that array to call the method to count duplicates inside the loop. Each word gets processed separately for each iteration!

[@abhishekanand3036]

@@rdnkmr will this work for string array ?

[@rdnkmr]

@@abhishekanand3036 when it is already a array you can directly loop the array and pass each string to that method

[@basirobaidi667]

First here 😊

[@nishikantwairkar6866]

❤️

[@bootcamprag]

How about you solve this using Python. I think you will use less code and complexity.

[@nagamanickam5702]

This is too complicated and verbose.. Simply use streams