Find the first repeating element in an array of integers

Given an array of integers, find the first repeating element in it. We need to find the element that occurs more than once and whose index of first occurrence is smallest.

Examples: 

Input:  arr[] = {10, 5, 3, 4, 3, 5, 6}
Output: 5 [5 is the first element that repeats]

Input:  arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10}
Output: 6 [6 is the first element that repeats]


A Simple Solution is to use two nested loops. The outer loop picks an element one by one, the inner loop checks whether the element is repeated or not. Once we find an element that repeats, we break the loops and print the element. Time Complexity of this solution is O(n2)

We can Use Sorting to solve the problem in O(nLogn) time. Following are detailed steps.
1) Copy the given array to an auxiliary array temp[].
2) Sort the temp array using a O(nLogn) time sorting algorithm.
3) Scan the input array from left to right. For every element, count its occurrences in temp[] using binary search. As soon as we find an element that occurs more than once, we return the element. This step can be done in O(nLogn) time.

We can Use Hashing to solve this in O(n) time on average. The idea is to traverse the given array from right to left and update the minimum index whenever we find an element that has been visited on right side. Thanks to Mohammad Shahid for suggesting this solution.

Following are C++ and Java implementation of this idea.

C++

/* C++ program to find first repeating element in arr[] */
#include<bits/stdc++.h>
using namespace std;
  
// This function prints the first repeating element in arr[]
void printFirstRepeating(int arr[], int n)
{
    // Initialize index of first repeating element
    int min = -1;
  
    // Creates an empty hashset
    set<int> myset;
  
    // Traverse the input array from right to left
    for (int i = n - 1; i >= 0; i--)
    {
        // If element is already in hash set, update min
        if (myset.find(arr[i]) != myset.end())
            min = i;
  
        else   // Else add element to hash set
            myset.insert(arr[i]);
    }
  
    // Print the result
    if (min != -1)
        cout << "The first repeating element is " << arr[min];
    else
        cout << "There are no repeating elements";
}
  
// Driver method to test above method
int main()
{
    int arr[] = {10, 5, 3, 4, 3, 5, 6};
  
    int n = sizeof(arr) / sizeof(arr[0]);
    printFirstRepeating(arr, n);
}
//This article is contributed by Chhavi

Java

/* Java program to find first repeating element in arr[] */
import java.util.*;
  
class Main
{
    // This function prints the first repeating element in arr[]
    static void printFirstRepeating(int arr[])
    {
        // Initialize index of first repeating element
        int min = -1;
  
        // Creates an empty hashset
        HashSet<Integer> set = new HashSet<>();
  
        // Traverse the input array from right to left
        for (int i=arr.length-1; i>=0; i--)
        {
            // If element is already in hash set, update min
            if (set.contains(arr[i]))
                min = i;
  
            else   // Else add element to hash set
                set.add(arr[i]);
        }
  
        // Print the result
        if (min != -1)
          System.out.println("The first repeating element is " + arr[min]);
        else
          System.out.println("There are no repeating elements");
    }
  
    // Driver method to test above method
    public static void main (String[] args) throws java.lang.Exception
    {
        int arr[] = {10, 5, 3, 4, 3, 5, 6};
        printFirstRepeating(arr);
    }
}

Output: 

The first repeating element is 5

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



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