Find pair number in an Array

Question

There is an array consisting several numbers. Among them find out the pair number(that consist 2 times in that array) in java. suppose {2,5,7,8,2,3,5,6,5} in this array 2 consist 2 times so its is pair number.

i tried this way :

HashMap<Integer, Integer> hmap = new HashMap<>();
  for (int i = 0; i < arr.length; i++)
    {
        Integer c = hmap.get(arr[i]);
        if (hmap.get(arr[i]) == null)
               hmap.put(arr[i], 1);
        else
          hmap.put(arr[i], ++c);
    }

sort the array, and for each element check if next element is same — kapilpatwa93, Nov 29 '16 at 06:47
put in map, keep infringement based on key, print all with counter =2 — Danyal Sandeelo, Nov 29 '16 at 06:48
[duplicate in an array](http://stackoverflow.com/questions/14944458/find-duplicate-element-in-array-in-time-on) — Aimee Borda, Nov 29 '16 at 06:49
Why ? Is it a requirement, a test, an exam ? What can you use ? And the last and most important .... **What have you tried ?** Please, see [ask] — AxelH, Nov 29 '16 at 06:49
@Kaustav, you are new so I would only tell you that you should take a [tour] to see of SO works. We are not going to give you code without you showing us a bit of concern. You need to tried this by yourself and show us what you have, and what is not working to guide you. This is not a free coding service ;) — AxelH, Nov 29 '16 at 06:51
i have tried using map .but didn't get the exact logic after iterating the array. — Kaustav, Nov 29 '16 at 06:56
@Kaustav Edit you question with the closest code you have had ... — AxelH, Nov 29 '16 at 07:00
That's a bit better, now explain what is the result with a specific input. And tell what is the problem. Also, you want pair or pairs ? — AxelH, Nov 29 '16 at 07:22
@AxelH i want to find out that number which occur 2 times in an array like pair number. — Kaustav, Nov 29 '16 at 07:24
Well, you already have ... your map contains the count off each values. Just need to print the result. Let see — AxelH, Nov 29 '16 at 07:26
I don't want to print all the values ,i want to print only pair numbers.i am looking for that the logic ?@AxelH — Kaustav, Nov 29 '16 at 07:30
@Kaustav Did you check my answer, this iterate on all the keySet (so the values in the initial array) and check the value (the number of representation). — AxelH, Nov 29 '16 at 07:36

score 2 · Answer 1 · answered Nov 29 '16 at 06:52

2

you can use streams

List<Integer> collect = Stream.of(2, 5, 7, 8, 2, 3, 5, 6, 5)
            .collect(Collectors.groupingBy(e -> e)).entrySet().stream()
            .filter(e -> e.getValue().size() == 2).map(Map.Entry::getKey)

answered Nov 29 '16 at 06:52

PawelN

848
7
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divine · Answer 2 · 2016-11-29T07:59:44.347

1

Program using primitive type

num[j]= '0'; // this is just a placeholder

  int []num = {2,5,7,8,2,3,5,6};
        int counter = 0;
        for(int i=0;i<num.length-1;i++){
            for(int j=i+1;j<num.length && num[i]!=0;j++){
                if(num[i] == num[j]){
                    counter++;
                    num[j]= 0;
                }
            }
            if(counter==1)
                 System.out.println("pair found for: "+num[i]);
            counter = 0;
        }

edited Nov 29 '16 at 07:59

answered Nov 29 '16 at 07:35

divine

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He already have a solution to find the pairs. And you should break the loop in some case (if num[i] = 0, then this is a checked value, skip it) – AxelH Nov 29 '16 at 07:44
o/p is : pair found for: 2 pair found for: 5 // 5 occurs 3 times in the array then how it can be pair? – Kaustav Nov 29 '16 at 07:51
@AxelH i opened the question at the beginning and i didnot find the hashmap code at that time. i forgot to refresh the browser. sorry. and i fixed the answer – divine Nov 29 '16 at 08:01

score 0 · Answer 3 · answered Nov 29 '16 at 06:54

0

I can think of 2 ways. First you could use a map, in this method you would take each element in an array and put it inside a map so that each time you iterate you could check if you already insert that value into the map.

The second method would be similar to a sort, but instead you could check each element in the array with all the following elements and as soon as you find a match you would return and exit. This method would be slower than the map method I think but faster than sorting it first and then finding adjacent values.

answered Nov 29 '16 at 06:54

George Miranda

150
1
13

Also this is assuming there is only 1 pair, since you said "find out the pair" – George Miranda Nov 29 '16 at 06:59
`check each element in the array with all the following elements and as soon as you find a match you would return and exit` How would you know that the value in position 2 is not the equivalent of the value in position 0, already checked ? EDIT : Yep, your comment kind of answer my question ^^ – AxelH Nov 29 '16 at 07:02

score 0 · Answer 4 · answered Nov 29 '16 at 06:57

Not posting the actual code, put that in Map and look for count. But you need to be careful to check for counting of 2s. Its not just 2, all counts with 4,6,8 etc also would lead to pairs(I think this point is lacking in all the answers and comments so far for this question) .

score 0 · Answer 5 · answered Nov 29 '16 at 07:28

0

From what you have posted, your code is good. You have a map that contains the count for each values found.

Now, you just need to print what you want,

for(Integer i : hmap.keySet()){
    Integer cnt = hmap.get(i);
    if(cnt > 1)} //or == 2
        System.out.println(i + " -> " + cnt);
    }
}

And you have an nice little input for each pair (with the count if there is more), just update your output.

answered Nov 29 '16 at 07:28

AxelH

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Yah, now i got the logic and its working fine for me. Thanks a lot :) – Kaustav Nov 29 '16 at 07:46
@Kaustav You are welcome, in the futur, think about giving every information from the start ;) ;) If you found a answer that match you needs, don't hesitate to accept it (just belove the vote system of the answer) to close the question. And upvote every answers that you found usefull. – AxelH Nov 29 '16 at 07:51
I will .Thanks :) – Kaustav Nov 29 '16 at 07:53
@Kaustav I was just reading your other question (although it was deleted) and reviewing your activity, I noticed you have still not marked this answer as Accepted. – Patrick Parker Dec 05 '16 at 16:53

Find pair number in an Array

5 Answers5