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Table of contents
1.
Introduction 
2.
Approach: Using Hashing
2.1.
Approach: Use unordered_map for hashing
2.2.
Implementation in Java
2.3.
Output 
2.4.
Implementation in C++
2.5.
Output 
2.6.
Implementation in Python3
2.7.
Output
2.8.
Implementation in C#
2.9.
Output 
3.
Frequently Asked Questions
3.1.
In an array, how do you find repeated elements?
3.2.
How do you use a for loop to locate duplicates in an array?
3.3.
Is it true that the spread operator removes duplicates?
3.4.
How do you search an array for unique elements?
3.5.
Duplicates in an array are removed in what way?
4.
Conclusion
Last Updated: Mar 27, 2024
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Find duplicates in a given array when elements are not limited to a range

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Introduction 

In this article, we will  find the duplicate number in a limited range array of size n with elements between 1 and n-1. 

Now let us understand the concept of finding duplicates with the help of an example.

For example,

Input:  { 1, 7, 8, 2, 7, 8}

Output: The duplicate element in the array is 7, 8

Input:  { 5, 8, 9, 2, 9, 5,3,6}

Output: The duplicate element in the array is 9,  5

 

Approach: Using Hashing

As we know, a data structure that associates values with keys is known as a hash table. This calculates key indexes using a hash function. We can store the value at the proper location based on the Hash Table index.

Now that you've devised a concept, let's look at its algorithm and implementation:

Approach: Use unordered_map for hashing

Count the number of times each element appears, and the elements with a frequency greater than one are printed. Because the range of numbers is unknown, unordered_map is used. 

Implementation in Java

/*Java program to find duplicates in the array given*/
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
 
public class Main
{
    // Driver program
    public static void main(String[] arg)
    {
        int array[] = {11, 10, 40, 11, 7, 9, 5, 40, 11};
        int n = array.length;
        printingDuplicates(array, n);
    }
    //finding and printing duplicates
    private static void printingDuplicates(int[] array, int n)
    {
        Map<Integer,Integer> map = new HashMap<>();
        int count = 0;
        boolean dupl = false;
        for(int i = 0; i < n; i++){
            if(map.containsKey(array[i])){
                count = map.get(array[i]);
                map.put(array[i], count + 1);
            }
            else{
                map.put(array[i], 1);
            }
        }
         
        for(Entry<Integer,Integer> entry : map.entrySet())
        {
            /* if the frequency is greater than 1
            printing the element */
            if(entry.getValue() > 1){
                System.out.print(entry.getKey()+ " ");
                dupl = true;
            }
        }
        // if no duplicates are present
        if(!dupl){
            System.out.println("-1");
        }
    }
}

Output 

Implementation in C++

#include <bits/stdc++.h>
using namespace std;
 
// function duplicates to find and print
void Duplicates(int arry[], int n)
{
    // using unordered_map to store frequencies
    unordered_map<int, int> freqc;
    for (int i=0; i<n; i++)
        freqc[arry[i]]++;
 
    bool dupl = false;
    unordered_map<int, int>:: iterator itr;
    for (itr=freqc.begin(); itr!=freqc.end(); itr++)
    {
        // if the frequency is more than 1 print element
        if (itr->second > 1)
        {
            cout << itr->first << " ";
            dupl = true;
        }
    }
 
    //If duplicates are not present
    if (dupl == false)
        cout << "-1";
}
 
// to test the driver program
int main()
{
    int arry[] = {7, 18, 10, 10, 7, 6, 11, 12, 11};
    int n = sizeof(arry) / sizeof(arry[0]);
    Duplicates(arry, n);
    return 0;
}

Output 

Implementation in Python3

To store a number as a key and its frequency as a value in Python, use a dictionary. Because the range of integers is unknown, a dictionary can be used.

# Python3 program for finding duplicates
# by using a dictionary approach.
def printingDuplicates(arry):
    dict = {}
 
    for elem in arry:
        try:
            dict[elem] += 1
        except:
            dict[elem] = 1
             
    for itm in dict:
         
         # if the frequency is greater than 1
         # then printing  the element
        if(dict[itm] > 1):
            print(itm, end=" ")
 
    print("\n")
 
# Driver Code
if __name__ == "__main__":
    list = [13, 13, 11, 6,
            2, 6, 5, 12, 11]
    printingDuplicates(list)

Output

 

Implementation in C#

/* C# program for finding the duplicates in given array*/
using System;
using System.Collections.Generic;
 
class findingDuplicates
{
    // function for finding and printing duplicates
    static void printingDuplicates(int[] arry, int n)
    {
        Dictionary<int,
                   int> map = new Dictionary<int,
                                             int>();
        int count = 0;
        bool dupl = false;
        for (int i = 0; i < n; i++)
        {
            if (map.ContainsKey(arry[i]))
            {
                count = map[arry[i]];
                map[arry[i]]++;
            }
            else
                map.Add(arry[i], 1);
        }
 
        foreach (KeyValuePair<int,
                              int> entry in map)
        {
            // if the frequency is greater than 1
            // printing the element
            if (entry.Value > 1)
                Console.Write(entry.Key + " ");
            dupl = true;
        }
 
        // if no duplicates found
        if (!dupl)
            Console.WriteLine("-1");
    }
 
    // Driver Code
    public static void Main(String[] arg)
    {
        int[] arry = { 3, 4, 4, 7,
                     5, 4, 5, 1, 1 };
        int n = arry.Length;
        printingDuplicates(arry, n);
    }
}

Output 

The above solution has time complexity O(N) and auxiliary space is  O(N)

You can also read about the Longest Consecutive Sequence.

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Frequently Asked Questions

In an array, how do you find repeated elements?

To find the repeated elements you first need to make a new array and initialise it with data.The we need two loops to find duplicate elements, where the outer loop iterates across the array from 0 to the array's length. If a match is detected, it indicates the duplicate element has been found, so display it.

How do you use a for loop to locate duplicates in an array?

In brute force approach for loop will be used to sort and iterate duplicates will be adjacent to each other and easy to detect after the array has been sorted.In a HashSet approach, we can populate this while iterating & abort if a match is found.As the for loop will run for n time it will have O(n) time and O(n) space complexity on average.

Is it true that the spread operator removes duplicates?

Remove duplicate elements and sort an array of numbers and objects using the spread operator without destroying the original array.

How do you search an array for unique elements?

There are several ways to find unique elements from an array such as by storing all of the elements in the key of the hashmap, using the nested loop technique or by sorting is an effective method and by the use of hashing.

Duplicates in an array are removed in what way?

There are two ways to remove duplicate elements from an array. One of it is using a temporary array or using a distinct index. The array must be in sorted order in order to delete the duplicate element. If the array is not sorted, use the Arrays. sort(arr) method to sort it.

Conclusion

In this blog we have discussed approaches in C++, Java, Python3 and C# using unordered hashmap ,to find duplicates in a given array when elements are not limited to a range. You can also look at more exciting blogs and articles curated by the coding ninja's team by visiting the Library.

We hope that this blog has helped you enhance your knowledge regarding finding duplicates in arrays in C++ and if you would like to learn more, check out our articles on Arrays in C++.

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