How to Find Duplicates Elements in Java Array? - Java Interview Questions -5
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- เผยแพร่เมื่อ 22 พ.ย. 2017
- How to Find Duplicates Elements in Java Array: The most important interview question.
Solution 1 : with Time Complexity = O(nxn)
Our first solution is very simple. All we are doing here is to loop over an array and comparing each element to every other element. Since we are comparing every element to every other element, this solution has quadratic time complexity i.e. O(n^2). This solution has the worst complexity in all three solutions.
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Solution 2 : with Time Complexity = O(n)
Second solution is even simpler than this. All you need to know is that Set doesn't allow duplicates in Java. Which means if you have added an element into Set and trying to insert duplicate element again, it will not be allowed. If add() returns false it means that element is not allowed in the Set and that is your duplicate.
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Solution 3 : with Time Complexity = O(2n)
Third solution takes advantage of another useful data structure, hash map. All you need to do is loop through the array using enhanced for loop and insert each element and its count into hash table.
In order to build map, you check if hash table contains the elements or not, if it is then increment the count otherwise insert element with count 1. Once you have this map ready, you can iterate over hashmap and print all those keys which has values greater than one. These are your duplicate elements. This is in fact a very good solution because you can extend it to found count of duplicates as well.
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شكرا للمشاهدة - วิทยาศาสตร์และเทคโนโลยี
King of the explanation, very clear and up to point. God Bless you!
Your teaching is apple of my eye
Your videos are really awesome, never saw anyone explaining things at this level. Please carry on for other concepts too.
I had this question in 3 consecutive interviews recently. But the only difference is they asked me to find duplicate and non-duplicate characters from a given string such as "ascedcsf". Your video is really very well organized and well explained.
Hi Naveen, Thanks for the amazing videos. I am preparing for Selenium with Java interview and referring your videos and I must say the way you explain, the concepts becomes very easy to understand for a JAVA naive like me as well.
One request, please increase the font size as its difficult to read your code in such small font. Thanks.....
There is one more question where we need to give the number of times a specific element is present i.e. for below array
String nameArray[] = {"Java","C#","Ruby","Python","Java","C++","C#","AngularJS","Java"};
it should return 3 for Java and 2 for C#.
For this we just need to add one extra line in HaspMap code:
System.out.println("The occurence of element "+entry.getKey()+" is "+ entry.getValue());
Full code below:
String nameArray[] = {"Java","C#","Ruby","Python","Java","C++","C#","AngularJS","Java"};
Map map = new HashMap();
for(String name: nameArray)
{
Integer count = map.get(name); // it will return null as no values are present for the key
if(count==null)
{
map.put(name, 1); //when its null add the value
}
else
{
map.put(name, ++count); // when it finds second occurrence of Java (i.e count =1 )we increase the count
}
}
System.out.println(map);
for(Map.Entry entry:map.entrySet())
{
if(entry.getValue()>1)
{
System.out.println("The duplicate element is: "+entry.getKey());
System.out.println("The occurence of element "+entry.getKey()+" is "+ entry.getValue());
}
}
String names[]={"Ankur","Hero","Arun","Ankur","Hero"};
HashMap map = new HashMap();
for (String ch : names) {
if (map.containsKey(ch)) {
int val = map.get(ch);
map.put(ch, val + 1);
} else {
map.put(ch, 1);
}
}
System.out.println(map);
}
}
String names[]= {"Java","javaScript","Ruby","C","Python","Java","C"};
Map storeMap=new HashMap();
for(String name:names){
Integer count= storeMap.put(name, 1);
if(count!=null){
System.out.println("Duplicate element is:::"+name);}}
Thanks Naveen. I understand each and every part because of this video.
awesome tutorial sir, thanks a lot for sharing, keep it up 👍👍🙂🙂
Thanks Naveen. That was crystal clear!
Amazing broad concepts !
one of the best solutions i have seen, it was so easy to understand the concepts. Thanks Naveen Sir:)
Thank you so much Naveen.. Really helpful..
Thanks Naveen, have been struggling a lot to get this understanding, thanks a ton! 😊😊
Hi Naveen, your explanation of the concepts are too good. I have learnt so many things while going thru your videos. One suggestion I would like to let you know that please increase font so it will be more visible when we watch your trainings in mobile as well.
Thanks,
Bharath.
Nice video concept is so clear
you are the best; i appreciate the work and time,
Hey Naveen, Another option is to use keySet() to get the getValue()
Really good teaching
Your genius naveen
Thank you Naveen.
One of the best explanation I found.Thanks a lot, for sharing.
w.a p to find the words number of time repeating using hashmap (string="web internet web chrome internet safari") how can i solve please tell me frds or any video links share me frds last intreviewer
ask me
Very well explain naveen thank you. I understand each & every part because of this video. I need optimize solution for few assignment like longest substring of non repeating character, intersecting of rectangles, no repeating first character in array, expression validation in java. If you give any idea it would help me lot .
excellent video sir.Please make more videos regd Java
The third solution can be optimized
If(hashmap.get(name) == null){
Hashmap.put(name,counter)
}
else {
Print duplicate element(name)
}
In this case we need not go for entryset
not optimised. if you are going to print like that, duplicate elements will printed many times
Hi Navee,n for HashMap in the else statement if i print the item or simple return that item do i really need to iterate over hashMap to find the count having value greater than 1.
Aap kaafi achha padha rahe ho sir
Thanks naveen, also could you please make same thing using recursive function. It will help us.
sir, please help me i really confuse between them what is the difference between the "array.length and array.length()" some time array.length will work and some time array.length() will work. is there is any rules for is for using it? with String arr.length() will work but when we use an integer type, its work with array.length.
If i have 5 billions elements in java array along with duplicates then also is this the best solution.Because i have been asked this question in interview but i couldn't give them answer.
Hey Naveen,
thanks for uploading such type of program but i have one inside (store.addname==false)
Set s=new HashSet();
for (String name : names) {
if(store.add(name)==false){
Thank you.
thank you!!
What if we sort the array and then compare only contiguous cells /indeces? That might be quicker.
We can iterate hash map like this
for(Map.Entry entrySet : map.entrySet())
{
if((int)entrySet.getValue()>1)
System.out.println("Duplicate element in an array:"+ (String)entrySet.getKey());
}
Navin, need help can you suggest solution for finding int duplicate from scanner
Naveen Thanks for sharing video but your logic using Hashset will not work if we will have more than two elements having same vaues in an array. Can You check that case?
Hi Naveen, Can we use Treeset instead of hashset?
In those concepts which is efficient way to use.....so plzz reply
Hey everyone, can anyone help me to find answer to this question? When we make use of in-built Java instructions in the video to check if the element is already in a Set, or HashMap, .... (e.g. , if (map.containsKey(ch))......) does Java do an additional iteration internally (abstract to the programmer) to check the whole set or hash map for the duplicates and therefore adding to the time complexity of the whole code?
100% they will ask you, in each video on selenium and java you are telling this ))
yea because whatever he is teaching everything asked in interview atleast i got several interview questions from his videos
Hi Naveen , If we have n same element , it will print n-1 times the same element for the set solution you provided.For example {harsh,raj, harsh,kumar,harsh} . output will be harsh, harsh
I like Hashset :)
Using HashMap also we can make it simple. but when we need to get the total count of duplicate we can modify the value fi key exist .
thanks amzing videao ..one request could you pls increase the fonts of eclipse..would be a great while watching
Go to Google
which ide u used in this video?
I seen many channels but not standard one like this
Thanks for watching
Tqs for the reply sir
Why here Object created with Parent Interface Reference ?
We can also create the object as per below whats the problem here ?
HashMap storeMap = new HashMap() instead of Map storeMap = new HashMap();
HashSet store = new HashSet() instead of Set store = new HashSet();
can someone explain this please ? Thanks in Advance.
please add printing pyramid pattern in java interview question
for String s = {"Java", "JavaScript", "Ruby", "Python", "Java", "C", "C", "Java"}, The first solution will not work. It will show "Java" Three times. Anyways, I would like to thank you for making these great tutorials.
super explaination please zoom the code while you explaining for better view..sometimes its not clear view
The code is clearly visible, you can change . you can change quality to 720p or 1080p
Hey Naveen u can do video for How to Find Duplicates Elements in Normal Int arrays?
Do you have series of Data Structures ?
Best solution is HashMap because 1st(Bruteforce) and 2nd(Set) can only compare 2 elements.But using HashMap we can find N no. of duplicate elements
can you provide the solution with HashMap?
@@Nancy-xd7bp here you go:String a1[]= {"java","php","python","java","c","c#","php","java"};
HashMap map=new HashMap();
for(String myNames : a1)
{
if(map.containsKey(myNames))
{
int count=map.get(myNames);
map.put(myNames, count+1);
}
else
{
map.put(myNames, 1);
}
}
System.out.println(map);
@@priteshpatel4312 superb solution...thank you
By using Hashset ---> This will print Java 2 times , if the there more than 2 occurrence of the same element . {"Java","Java","Java"}
Hi how internally set remove duplicate in o(n) can you explain it
Which add method call and how it will remove any object which is duplicate.
Set by default stores only unique value in it. So here add(value) checks if the particular value is stored in it or not. If it finds the value then it will not add it in the set otherwise it will add. Let me know if i am clear.
@@sayanbhaumik5739 how it will verify weather it object is present previously if bean object doesn't override hashcode and equals method
Can you explain why the j value is not starting from 0 why it's started j=I+1;
you would not want to compare any element with itself, otherwise all the elements will be printed as duplicate. Its j=i+1 because you want to compare with next element onward.
String[] names = {"java","phython","java","c#","phython",".net"};
Arrays.stream(names)
.filter(name -> Collections.frequency(Arrays.asList(names), name) > 1)
.collect(Collectors.toSet())
.forEach(System.out::println);
Hi naveen, i could not find a solution to check for duplicates in this array: integer array [12,12,7,7,7,7,7] , can you please help?
public static void main(String[] args) {
int[] arr = {1, 5, 12, 12,7,7,7,7};
Set uniqueSet = new HashSet();
Set duplicates = new HashSet();
for (int num : arr) {
if (!uniqueSet.add(num)) {
duplicates.add(num);
}
}
if (!duplicates.isEmpty()) {
System.out.println("Duplicate elements in the array without repetition:");
for (int duplicate : duplicates) {
System.out.println(duplicate);
}
} else {
System.out.println("There are no duplicate elements in the array");
}
}
Hi Naveen,
thanks for sharing.iam getting output as shown below, how can i get only one value, can u solve please
String names[]= {"java","phython","java","c#","phython",".net"};
Map storemap=new HashMap();
for(String name : names) {
Integer count=storemap.get(name);
if(count==null) {
storemap.put(name, 1);
}else {
storemap.put(name,++count);
}
//get the value from this hashmap
Set entryset= storemap.entrySet();
for(Entry entry : entryset) {
if (entry.getValue()>1) {
System.out.println("Duplicate value is :"+entry.getKey());
}
}
}
output:
Duplicate value is :java
Duplicate value is :java
Duplicate value is :java
Duplicate value is :phython
Duplicate value is :java
Duplicate value is :phython
Hi Srividya,
End the first for loop before getting the value from HashMap.Hope this will help
String names[]= {"java","phython","java","c#","phython",".net"};
Map storemap=new HashMap();
for(String name : names) {
Integer count=storemap.get(name);
if(count==null) {
storemap.put(name, 1);
}else {
storemap.put(name,++count);
}
//get the value from this hashmap
}
Set entryset= storemap.entrySet();
for(Entry entry : entryset) {
if (entry.getValue()>1) {
System.out.println(entry.getValue()+" times string "+entry.getKey()+"duplicated");
}
}
Using HashMap also we can make it simple. I am checking for duplicate numbers so code is written with int. Like this...
int a[]= {2,2,4,6,7,6,9,9};
HashMap map = new HashMap();
for(Integer num1:a) {
if(map.put(num1, num1)!=null) {
System.out.println("Duplicate using Hashmap is: "+ num1);
}
}
thanks brother!
pls use simple for loops
in first example (int i=0 ; i
I think there is the bug in this code I am not sure , but let assume String a[] = {"Java", "Python", "C", "C#", "C++", "Go", "C", "C", "Java", "Java"}; how this should with your code .
How to find non duplicates in java please help me
what is time complexity?
The amount of time it takes for completion.
String[] dupArray={"Mumbai","Chennai","Delhi","Kochi","Chennai","Kochi","Trivandrum","Delhi"};
Map myMap=new HashMap();
int counter=0;
for (String city:dupArray) {
if(myMap.put(city,++counter)!=null){
System.out.println("Duplicate Value="+city);
}
}
write a javascript code to perform the following on n array N containing the following{"welcome", "friends"} (i)Add "try" to array N.(ii)Join array N to another array M. (iii)Delete the last element from the array M. (iv)sort the array (v)Display the array in reverse. pls solve this.. if u're a real teacher
String names[]= {"Java","javaScript","Ruby","C","Python","Java","C"};
Map storeMap=new HashMap();
for(String name:names){
Integer count= storeMap.put(name, 1);
if(count!=null){
System.out.println("Duplicate element is:::"+name);}}