Java + Count duplicates from int array without using any Collection or another intermediate Array

Question

As a part of the Java interview question paper I have got following issue to solve. But I am bit wonder whether how can I implement it without any Collection or intermediate

Answer 1

Keeping one extra variable for maintaining count, plus sorting of array in the initial phase.

public static void main(String[] args) {
        int[] numbers = { 7, 2, 6, 1, 4, 7, 4, 5, 4, 7, 7, 3, 1 };
        Arrays.sort(numbers);
        System.out.println("Sorted Array is :: = " + Arrays.toString(numbers));

        int count = 0;
        int tempCount = 0; // to keep local count of matched numbers
        String duplicates = "";
        for (int i = 1; i < numbers.length; i++) {
            if (numbers[i] == numbers[i - 1]) {
                if ((tempCount == 0)) { // If same number is repeated more than
                                        // two times, like 444, 7777
                    count = count + 1;
                    tempCount = tempCount + 1;
                    duplicates = duplicates.concat(Integer.toString(numbers[i])
                            + ",");
                }
            } else {
                tempCount = 0;
            }
        }

        System.out.println("No of duplicates :: = " + count);
        System.out.println("Duplicate Numbers are :: = " + duplicates);
    }

output

Sorted Array is :: = [1, 1, 2, 3, 4, 4, 4, 5, 6, 7, 7, 7, 7]
No of duplicates :: = 3
Duplicate Numbers are :: = 1,4,7,

Answer 2

This is the simplest solution I can think of. I just added an extra counter so that integers with two or more repetitions still in the array are ignored.

static int findNumber(int[] arr) 
{  
    int duplicateCounter = 0;

    System.out.print("Duplicates: ");

    for(int i = 0; i < arr.length; i++)
    {
        boolean duplicate = false;
        int numOfOccurrences = 1;

        for (int j = (i+1); j < arr.length; j++)
        {
            if (arr[i] == arr[j])
            {
                numOfOccurrences++;
                duplicate = true;
            }
        }
        if(numOfOccurrences == 2 && duplicate == true)
        {
            duplicateCounter++;
            System.out.print(arr[i] + " ");
        }
    }

    return duplicateCounter;
}

My test run: Test run

Input: 1, 2, 3, 4, 2, 4, 1, 1, 1

Duplicates: 2 4 1

Number of duplicates: 3

Answer 3

Below method not use any collection, just use Arrays.sort() method to help sort array into ascending order as default, e.g array = [9,3,9,3,9] will sort into [3,3,9,9,9].If input [9,9,9,9,9], expected result is 1, since only repeated number is 9.If input [9,3,9,3,9,255,255,1], expected result is 3, since repeated numbers are 3,9,255. If input [7,2,6,1,4,7,4,5,4,7,7,3,1], expected result is 3, since repeated numbers are 1,4,7.

public static int findDuplicateCountsInArray(int[] nums) {
    // Sort the input array into default ascending order
    Arrays.sort(nums);
    int prev = nums[0];
    int count = 0;
    // Recording a number already a repeated one
    // e.g [9,9,9] the 3rd 9 will not increase duplicate count again
    boolean numAlreadyRepeated = false;
    for(int i = 1; i < nums.length; i++) {
        if(prev == nums[i] && !numAlreadyRepeated) {
            count++;
            numAlreadyRepeated = true;
        } else if(prev != nums[i]) {
            prev = nums[i];
            numAlreadyRepeated = false;
        }
    }
    return count;
}

Answer 4

Here I have written code in JAVA. also the inputted numbers, have been considered as String. This question has also been added to CODEWARS. and I hope this simple solution helps You

public class countingduplicates {
  public static void main(String[] args) {
    int i=0,j=0,c=0,a=0;
    String text="7261474547731";
    text=text.toLowerCase();

    for(i=0; i<text.length(); i++) {
      for(j=0; j<text.length(); j++) {
        if(text.charAt(i) == text.charAt(j)) { 
          c++;
        }
      }

      System.out.println(text.charAt(i) + " occured " + c + " times");
      if(c>1) {
        a++;
      }  

      String d = String.valueOf(text.charAt(i)).trim();
      text = text.replaceAll(d,"");
      c = 0;
      i = 0; //cause i have trimmed the string and by default i increases by 1, so i have to assign it =0
      j = 0; //cause i have trimmed the string and by default j increases by 1, so i have to assign it =0
    }
  System.out.println("Total count of Duplicates:" + a);
  }
}