Counting repeated elements in an integer array

Question

I have an integer array crr_array and I want to count elements, which occur repeatedly. First, I read the size of the array and initialize it with numbers read from the console. In the array new_array , I store the elements that are repeated. The array times stores the number of consecutive occurrences of an element. Then, I try to search for the repeating sequences and print them in a specific format. However, it does not work.

// Get integer array size
Scanner input = new Scanner(System.in);
System.out.println("Enter array size: ");
int size = input.nextInt();

int[] crr_array = new int[size];
int[] new_array= new int[size];
int[] times = new int[size];

// Read integers from the console
System.out.println("Enter array elements: ");
for (int i = 0; i < crr_array.length; i++) {
    crr_array[i] = input.nextInt();
    times[i] = 1;
}

// Search for repeated elements
for (int j = 0; j < crr_array.length; j++) {
    for (int i = j; i < crr_array.length; i++) {
        if (crr_array[j] == crr_array[i] && j != i) {
            new_array[i] = crr_array[i];
            times[i]++;
        }
    }
}



//Printing output
for (int i = 0; i <  new_array.length; i++) {
    System.out.println("\t" + crr_array[i] + "\t" +  new_array[i] + "\t" + times[i]);

}

I want the output to look like this:

There are <count_of_repeated_element_sequences> repeated numbers 
<repeated_element>: <count> times
...

For example:

There are 3 repeated numbers:
22: 2 times
4: 3 times
1: 2 times

How can I find the repeated elements and their counts? How can I print them as shown above?

Answer 1

This kind of problems can be easy solved by dictionaries (HashMap in Java).

  // The solution itself 
  HashMap<Integer, Integer> repetitions = new HashMap<Integer, Integer>();

  for (int i = 0; i < crr_array.length; ++i) {
      int item = crr_array[i];

      if (repetitions.containsKey(item))
          repetitions.put(item, repetitions.get(item) + 1);
      else
          repetitions.put(item, 1);
  }

  // Now let's print the repetitions out
  StringBuilder sb = new StringBuilder();

  int overAllCount = 0;

  for (Map.Entry<Integer, Integer> e : repetitions.entrySet()) {
      if (e.getValue() > 1) {
          overAllCount += 1;

          sb.append("\n");
          sb.append(e.getKey());
          sb.append(": ");
          sb.append(e.getValue());
          sb.append(" times");
      }
  }

  if (overAllCount > 0) {
      sb.insert(0, " repeated numbers:");
      sb.insert(0, overAllCount);
      sb.insert(0, "There are ");
  }

  System.out.print(sb.toString());

Answer 2

If you have values in a short set of possible values then you can use something like Counting Sort

If not you have to use another data structure like a Dictionary, in java a Map

int[] array
Map<Integer, Integer> 

where Key = array value for example array[i] and value = a counter

Example:

int[] array = new int [50];
Map<Integer,Integer> counterMap = new HashMap<>();

//fill the array

    for(int i=0;i<array.length;i++){
         if(counterMap.containsKey(array[i])){
          counterMap.put(array[i], counterMap.get(array[i])+1 );
         }else{
          counterMap.put(array[i], 1);
         }
    }

Answer 3

public class DuplicationNoInArray {

    /**
     * @param args
     *            the command line arguments
     */
    public static void main(String[] args) throws Exception {
        int[] arr = { 1, 2, 3, 4, 5, 1, 2, 8 };
        int[] result = new int[10];
        int counter = 0, count = 0;
        for (int i = 0; i < arr.length; i++) {
            boolean isDistinct = false;
            for (int j = 0; j < i; j++) {
                if (arr[i] == arr[j]) {
                    isDistinct = true;
                    break;
                }
            }
            if (!isDistinct) {
                result[counter++] = arr[i];
            }
        }
        for (int i = 0; i < counter; i++) {
            count = 0;
            for (int j = 0; j < arr.length; j++) {
                if (result[i] == arr[j]) {
                    count++;
                }

            }
            System.out.println(result[i] + " = " + count);

        }
    }
}

Answer 4

for (int i = 0; i < x.length; i++) {

    for (int j = i + 1; j < x.length; j++) {

        if (x[i] == x[j]) {
            y[i] = x[i];
            times[i]++;
        }

    }

}

Answer 5

with O(n log(n))

int[] arr1; // your given array
int[] arr2 = new int[arr1.length];
Arrays.sort(arr1);

for (int i = 0; i < arr1.length; i++) {
    arr2[i]++;
    if (i+1 < arr1.length) 
    {
        if (arr1[i] == arr1[i + 1]) {
            arr2[i]++;
            i++;
        }
    }
}

for (int i = 0; i < arr1.length; i++) {
    if(arr2[i]>0)
    System.out.println(arr1[i] + ":" + arr2[i]);
}

Answer 6

You have to use or read about associative arrays, or maps,..etc. Storing the the number of occurrences of the repeated elements in array, and holding another array for the repeated elements themselves, don't make much sense.

Your problem in your code is in the inner loop

 for (int j = i + 1; j < x.length; j++) {

        if (x[i] == x[j]) {
            y[i] = x[i];
            times[i]++;
        }

    }

Answer 7

package jaa.stu.com.wordgame;

/**
 * Created by AnandG on 3/14/2016.
 */
public final class NumberMath {
    public static boolean isContainDistinct(int[] arr) {

        boolean isDistinct = true;
        for (int i = 0; i < arr.length; i++)

        {

            for (int j = 0; j < arr.length; j++) {
                if (arr[i] == arr[j] && i!=j) {
                    isDistinct = false;
                    break;
                }
            }

        }
        return isDistinct;
    }
    public static boolean isContainDistinct(float[] arr) {

        boolean isDistinct = true;
        for (int i = 0; i < arr.length; i++)

        {

            for (int j = 0; j < arr.length; j++) {
                if (arr[i] == arr[j] && i!=j) {
                    isDistinct = false;
                    break;
                }
            }

        }
        return isDistinct;
    }
    public static boolean isContainDistinct(char[] arr) {

        boolean isDistinct = true;
        for (int i = 0; i < arr.length; i++)

        {

            for (int j = 0; j < arr.length; j++) {
                if (arr[i] == arr[j] && i!=j) {
                    isDistinct = false;
                    break;
                }
            }

        }
        return isDistinct;
    }
    public static boolean isContainDistinct(String[] arr) {

        boolean isDistinct = true;
        for (int i = 0; i < arr.length; i++)

        {

            for (int j = 0; j < arr.length; j++) {
                if (arr[i] == arr[j] && i!=j) {
                    isDistinct = false;
                    break;
                }
            }

        }
        return isDistinct;
    }
    public static int[] NumberofRepeat(int[] arr) {

        int[] repCount= new int[arr.length];
        for (int i = 0; i < arr.length; i++)

        {

            for (int j = 0; j < arr.length; j++) {
                if (arr[i] == arr[j] ) {
                    repCount[i]+=1;
                }
            }

        }
        return repCount;
    }
}


call  by NumberMath.isContainDistinct(array) for find is it contains repeat or not

call by int[] repeat=NumberMath.NumberofRepeat(array) for find repeat count. Each location contains how many repeat corresponding value of array...

Answer 8

public static void main(String[] args) {
    Scanner input=new Scanner(System.in);
    int[] numbers=new int[5];
    String x=null;
    System.out.print("enter the number 10:"+"/n");
    for(int i=0;i<5;i++){
        numbers[i] = input.nextInt();
    }
    System.out.print("Numbers  :  count"+"\n");
    int count=1;
    Arrays.sort(numbers);
    for(int z=0;z<5;z++){
        for(int j=0;j<z;j++){
            if(numbers[z]==numbers[j] & j!=z){
                count=count+1;
            }
        }
        System.out.print(numbers[z]+" - "+count+"\n");
        count=1;

    }

Answer 9

public class ArrayDuplicate {
private static Scanner sc;
static int totalCount = 0;

    public static void main(String[] args) {
        int n, num;
        sc = new Scanner(System.in);
        System.out.print("Enter the size of array: ");
        n =sc.nextInt();
        int[] a = new int[n];
        for(int i=0;i<n;i++){
            System.out.print("Enter the element at position "+i+": ");
            num = sc.nextInt();
            a[enter image description here][1][i]=num;
        }
        System.out.print("Elements in array are: ");
        for(int i=0;i<a.length;i++)
            System.out.print(a[i]+" ");
        System.out.println();
        duplicate(a);
        System.out.println("There are "+totalCount+" repeated numbers:");
    }

    public static void duplicate(int[] a){
        int j = 0,count, recount, temp;
        for(int i=0; i<a.length;i++){
            count = 0;
            recount = 0;
            j=i+1;
            while(j<a.length){
                if(a[i]==a[j])
                    count++;
                j++;
            }
            if(count>0){
                temp = a[i];
                for(int x=0;x<i;x++){
                    if(a[x]==temp)
                        recount++;
                }
                if(recount==0){                 
                    totalCount++;
                    System.out.println(+a[i]+" : "+count+" times");
                }   
            }

        }
    }

}

Answer 10

package com.core_java;

import java.util.Arrays;
import java.util.Scanner;

public class Sim {
    public static void main(String[] args) {

        Scanner input = new Scanner(System.in);
        System.out.println("Enter array size: ");
        int size = input.nextInt();

        int[] array = new int[size];

        // Read integers from the console
        System.out.println("Enter array elements: ");
        for (int i = 0; i < array.length; i++) {
            array[i] = input.nextInt();
        }
        Sim s = new Sim();
        s.find(array);
    }

    public void find(int[] arr) {
        int count = 1;
        Arrays.sort(arr);

        for (int i = 0; i < arr.length; i++) {

            for (int j = i + 1; j < arr.length; j++) {
                if (arr[i] == arr[j]) {
                    count++;
                }
            }
            if (count > 1) {
                System.out.println();
                System.out.println("repeated element in array " + arr[i] + ": " + count + " time(s)");
                i = i + count - 1;
            }
            count = 1;
        }
    }

}

Answer 11

 public static void duplicatesInteger(int arr[]){
    Arrays.sort(arr);       
    int count=0;
    Set s=new HashSet();
    for(int i=0;i<=arr.length-1;i++){
        for(int j=i+1;j<=arr.length-1;j++){
            if(arr[i]==arr[j] && s.add(arr[i])){
                count=count+1;              
                                }
        }
        System.out.println(count);
    }
}

Answer 12

private static void getRepeatedNumbers() {

    int [] numArray = {2,5,3,8,1,2,8,3,3,1,5,7,8,12,134};
    Set<Integer> nums = new HashSet<Integer>();
    
    for (int i =0; i<numArray.length; i++) {
        if(nums.contains(numArray[i]))
            continue;
            int count =1;
            for (int j = i+1; j < numArray.length; j++) {
                if(numArray[i] == numArray[j]) {
                    count++;
                }
                    
            }
            System.out.println("The "+numArray[i]+ " is repeated "+count+" times.");
            nums.add(numArray[i]);
        }
    }
    

Answer 13

public class FindRepeatedNumbers 
{
 public static void main(String[] args) 
    {
     int num[]={1,3,2,4,1,2,4,6,7,5};
           Arrays.sort(num);

  for(int j=1;j<num.length;j++)
      {
       if(num[j]==num[j-1])
    {
            System.out.println(num[j]);

       }
   }

       }
     }