Array elements that appear more than once

Given an integer array, print all repeating elements (Elements that appear more than once) in the array. The output should contain elements according to their first occurrences.

Examples:

Input: arr[] = {12, 10, 9, 45, 2, 10, 10, 45}
Output: 10 45

Input: arr[] = {1, 2, 3, 4, 2, 5}
Output: 2

Input: arr[] = {1, 1, 1, 1, 1}
Output: 1

The idea is to use Hashing to solve this in O(n) time on average. We store elements and their counts in a hash table. After storing counts, we traverse input array again and print those elements whose counts are more than once. To make sure that every output element is printed only once, we set count as 0 after printing the element.  

Implementation:

C++

// C++ program to print all repeating elements
#include <bits/stdc++.h>
using namespace std;

void printRepeating(int arr[], int n)
{
    // Store elements and their counts in
    // hash table
    unordered_map<int, int> mp;
    for (int i = 0; i < n; i++)
        mp[arr[i]]++;

    // Since we want elements in same order,
    // we traverse array again and print
    // those elements that appear more than
    // once.
    for (int i = 0; i < n; i++) {
        if (mp[arr[i]] > 1) {
            cout << arr[i] << " ";

            // This is tricky, this is done
            // to make sure that the current
            // element is not printed again
            mp[arr[i]] = 0;
        }
    }
}

// Driver code
int main()
{
    int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printRepeating(arr, n);
    return 0;
}

Java

// Java program to print all repeating elements

import java.util.*;
import java.util.Map.Entry;
import java.io.*;
import java.lang.*;

public class GFG {

    static void printRepeating(int arr[], int n)
    {

        // Store elements and their counts in
        // hash table
        Map<Integer, Integer> map
            = new LinkedHashMap<Integer, Integer>();
        for (int i = 0; i < n; i++) {
            try {
                map.put(arr[i], map.get(arr[i]) + 1);
            }
            catch (Exception e) {
                map.put(arr[i], 1);
            }
        }

        // Since we want elements in the same order,
        // we traverse array again and print
        // those elements that appear more than once.

        for (Entry<Integer, Integer> e : map.entrySet()) {
            if (e.getValue() > 1) {
                System.out.print(e.getKey() + " ");
            }
        }
    }

    // Driver code
    public static void main(String[] args) throws IOException
    {
        int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
        int n = arr.length;
        printRepeating(arr, n);
    }
}

// This code is contributed by Wrick

Python3

# Python3 program to print
# all repeating elements
def printRepeating(arr, n):

    # Store elements and
    # their counts in
    # hash table
    mp = [0] * 100
    for i in range(0, n):
        mp[arr[i]] += 1

    # Since we want elements
    # in same order, we
    # traverse array again
    # and print those elements
    # that appear more than once.
    for i in range(0, n):
        if (mp[arr[i]] > 1):
            print(arr[i], end = " ")

            # This is tricky, this
            # is done to make sure
            # that the current element
            # is not printed again
            mp[arr[i]] = 0

# Driver code
arr = [12, 10, 9, 45,
       2, 10, 10, 45]
n = len(arr)
printRepeating(arr, n)

# This code is contributed
# by Smita

C#

// C# program to print all repeating elements
using System;
using System.Collections.Generic;

class GFG
{
static void printRepeating(int []arr, int n)
{

    // Store elements and their counts in
    // hash table
    Dictionary<int,
               int> map = new Dictionary<int,
                                         int>();
    for (int i = 0 ; i < n; i++)
    {
        if(map.ContainsKey(arr[i]))
        {
            var val = map[arr[i]];
            map.Remove(arr[i]);
            map.Add(arr[i], val + 1);
        }
        else
        {
            map.Add(arr[i], 1);
        }
    }

    // Since we want elements in the same order,
    // we traverse array again and print
    // those elements that appear more than once.
    foreach(KeyValuePair<int, int> e in map)
    {
        if (e.Value > 1)
        {
            Console.Write(e.Key + " ");
        }
    }
}

// Driver code
public static void Main(String[] args)
{
    int []arr = { 12, 10, 9, 45, 2, 10, 10, 45 };
    int n = arr.Length;
    printRepeating(arr, n);
}
}

// This code is contributed by PrinciRaj1992

Javascript

// JavaScript program to print all
// repeating elements

function printRepeating(arr, n)
{
    // Store elements and their counts in
    // hash table
    var mp = new Map();
    for (var i = 0; i < n; i++)
    {
        if(mp.has(arr[i]))
            mp.set(arr[i], mp.get(arr[i])+1)
        else
            mp.set(arr[i], 1)
    }

    // Since we want elements in same order,
    // we traverse array again and print
    // those elements that appear more than
    // once.
    for (var i = 0; i < n; i++) {
        if (mp.get(arr[i]) > 1) {
            console.log( arr[i] + " ");

            // This is tricky, this is done
            // to make sure that the current
            // element is not printed again
            mp.set(arr[i], 0);
        }
    }
}

// Driver code
var arr = [ 12, 10, 9, 45, 2, 10, 10, 45 ];
var n = arr.length;
printRepeating(arr, n);

Output

10 45 

complexity Analysis:

• Time Complexity: O(n) under the assumption that hash insert and search functions work in O(1) time.
• Auxiliary Space: O(n), where n represents the size of the given array.

Method #2:Using Built-in Python functions:

• Count all the frequencies of all elements using Counter() function.
• Traverse in this frequency dictionary and print all keys whose value is greater than 1.

Below is the implementation of above approach:

Python3

# Python3 program to print
# all repeating elements
from collections import Counter

def printRepeating(arr, n):

    # Counting frequencies
    freq = Counter(arr)

    # Traverse the freq dictionary and
    # print all the keys whose value
    # is greater than 1
    for i in freq:
        if(freq[i] > 1):
            print(i, end=" ")


# Driver code
arr = [12, 10, 9, 45,
       2, 10, 10, 45]
n = len(arr)
printRepeating(arr, n)

# This code is contributed by vikkycirus

C++

// C++ program to print
// all repeating elements

#include <iostream>
#include <unordered_map>

using namespace std;

void printRepeating(int arr[], int n)
{
    // Create an unordered_map to store the frequency of
    // each element
    unordered_map<int, int> freq;

    // Traverse the array and increment the count of each
    // element in the map
    for (int i = 0; i < n; i++) {
        freq[arr[i]]++;
    }

    // Traverse the map and print all the keys whose value
    // is greater than 1
    for (auto i : freq) {
        if (i.second > 1) {
            cout << i.first << " ";
        }
    }
}

int main()
{
    int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printRepeating(arr, n);
    return 0;
}

Java

import java.util.*;

public class Main {
    public static void printRepeating(int[] arr, int n)
    {
        // Counting frequencies
        Map<Integer, Integer> freq
            = new HashMap<Integer, Integer>();
        for (int i = 0; i < n; i++) {
            int key = arr[i];
            freq.put(key, freq.getOrDefault(key, 0) + 1);
        }

        // Traverse the freq dictionary and print all the
        // keys whose value is greater than 1
        for (int i : freq.keySet()) {
            if (freq.get(i) > 1) {
                System.out.print(i + " ");
            }
        }
    }

    public static void main(String[] args)
    {
        int[] arr = { 12, 10, 9, 45, 2, 10, 10, 45 };
        int n = arr.length;
        printRepeating(arr, n);
    }
}

Output

10 45 

complexity Analysis:

• Auxiliary Space: O(n), where n represents the size of the given array.

Method #3(Space Optimization) : we can use binary search lower_bound function to find first occurrence of arr[i] and Upper_bound function to find last occurrence of x and if the last_index-first_ind+1>1 means , arr[i] has more than one frequency.

Below is the implementation of above approach:

C++

// C++ implementation of the above approach

#include <bits/stdc++.h>
using namespace std;

//Function to Print elements that repeat more than 1
void printRepeating(int arr[], int n)
{   sort(arr,arr+n);//sort array for binary search

    for(int i=0;i <n ;i++)
    { //index of first and last occ of arr[i];
      int first_index = lower_bound(arr,arr+n,arr[i])- arr;
      int last_index = upper_bound(arr,arr+n,arr[i])- arr-1;

      int fre = last_index-first_index+1;//frequency of arr[i]

      if(fre > 1 )// elements that repeat more than 1
      {  i=last_index; //update i to last_index
       cout<<arr[i]<<" ";  }// print repeat element
    }
}

// Driver code
int main()
{
    int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
    int n = sizeof(arr) / sizeof(arr[0]);

    //Function call
    printRepeating(arr, n);
    return 0;
}

// This Approach is contributed by nikhilsainiofficial546

Java

import java.util.*;

class Main {
    // Function to Print elements that repeat more than 1
    public static void printRepeating(int arr[], int n)
    {
        Arrays.sort(arr);//sort array for binary search

        for (int i = 0; i < n; i++) {
            // index of first and last occ of arr[i];
            int first_index = Arrays.binarySearch(arr, arr[i]);
            int last_index = Arrays.binarySearch(arr, arr[i]);

            // to handle the case if the element is not found
            if (first_index < 0) {
                continue;
            }

            while ((last_index < n - 1) && (arr[last_index + 1] == arr[i])) {
                last_index++;
            }

            int fre = last_index - first_index + 1;//frequency of arr[i]

            if (fre > 1)// elements that repeat more than 1
            {
                i = last_index; // update i to last_index
                System.out.print(arr[i] + " "); // print repeat element
            }
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
        int n = arr.length;

        // Function call
        printRepeating(arr, n);
    }
}

Python3

# Python implementation
def printRepeating(arr, n):

    # Sort the array so that repeating
    # elements can be tracked
    arr.sort()

    # Initialize index = 0
    i = 0
    while i < n:
        first_index = i

        # Initialize last_index to i
        # plus the frequency
        last_index = i + arr.count(arr[i]) - 1

        # If frequency of current element
        # is more than 1, then print it
        if last_index - first_index > 0:
            print(arr[i], end = " ")

        # Update index to last index
        # plus 1
        i = last_index + 1

arr = [ 12, 10, 9, 45, 2, 10, 10, 45 ]
n = len(arr)

# Function call
printRepeating(arr, n)

# This code is contributed by akashish__

Javascript

function printRepeating(arr, n) {
  // Sort the array so that repeating elements can be tracked
  arr.sort();

  // Initialize index = 0
  let i = 0;
  while (i < n) {
    let first_index = i;

    // Initialize last_index to i plus the frequency
    let last_index = i + arr.filter(x => x === arr[i]).length - 1;

    // If frequency of current element is more than 1, then print it
    if (last_index - first_index > 0) {
      console.log(arr[i]);
    }

    // Update index to last index plus 1
    i = last_index + 1;
  }
}

let arr = [12, 10, 9, 45, 2, 10, 10, 45];
let n = arr.length;

// Function call
printRepeating(arr, n);

Output

10 45 

Time Complexity: O(n*log₂n), Take log₂n for binary search function(upper and lower bound)
Auxiliary Space: O(1)