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 is the 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); 
    } 
} 

Python3

# Python3 program to find first repeating 
# element in arr[]  
  
# This function prints the first repeating  
# element in arr[] 
def printFirstRepeating(arr, n): 
  
    # Initialize index of first repeating element 
    Min = -1
  
    # Creates an empty hashset 
    myset = dict() 
  
    # Traverse the input array from right to left 
    for i in range(n - 1, -1, -1): 
      
        # If element is already in hash set, 
        # update Min 
        if arr[i] in myset.keys(): 
            Min = i 
  
        else: # Else add element to hash set 
            myset[arr[i]] = 1
      
    # Print the result 
    if (Min != -1): 
        print("The first repeating element is",  
                                      arr[Min]) 
    else: 
        print("There are no repeating elements") 
  
# Driver Code 
arr = [10, 5, 3, 4, 3, 5, 6] 
  
n = len(arr) 
printFirstRepeating(arr, n) 
  
# This code is contributed by Mohit kumar 29 

C#

using System; 
using System.Collections.Generic; 
  
/* C# program to find first repeating element in arr[] */
  
public class GFG 
{ 
    // This function prints the first repeating element in arr[]  
    public static void printFirstRepeating(int[] arr) 
    { 
        // Initialize index of first repeating element  
        int min = -1; 
  
        // Creates an empty hashset  
        HashSet<int> set = new HashSet<int>(); 
  
        // 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) 
        { 
          Console.WriteLine("The first repeating element is " + arr[min]); 
        } 
        else
        { 
          Console.WriteLine("There are no repeating elements"); 
        } 
    } 
  
    // Driver method to test above method  
  
    public static void Main(string[] args) 
    { 
        int[] arr = new int[] {10, 5, 3, 4, 3, 5, 6}; 
        printFirstRepeating(arr); 
    } 
} 
  
// This code is contributed by Shrikant13 

Output

The first repeating element is 5

Another Approach:
If you want to do this without using any additional data structure. The problem can also be solved using array only. See the method below.

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) 
{ 
      
    // This will set k=1, if any  
    // repeating element found 
    int k = 0; 
  
    // max = maximum from (all elements & n) 
    int max = n; 
    for (int i = 0; i < n; i++) 
        if (max < arr[i]) 
            max = arr[i]; 
  
    // Array a is for storing  
    // 1st time occurence of element 
    // initialized by 0 
    int a[max + 1] = {}; 
  
    // Store 1 in array b  
    // if element is duplicate 
    // initialized by 0 
    int b[max + 1] = {}; 
  
    for (int i = 0; i < n; i++)  
    { 
      
        // Duplicate element found 
        if (a[arr[i]])  
        { 
            b[arr[i]] = 1; 
            k = 1; 
            continue; 
        } 
        else
            // storing 1st occurence of arr[i] 
            a[arr[i]] = i;  
    } 
  
    if (k == 0) 
        cout << "No repeating element found" << endl; 
    else 
    { 
        int min = max + 1; 
        
        // trace array a & find repeating element 
        // with min index 
        for (int i = 0; i < max + 1; i++)  
            if (a[i] && min > a[i] && b[i]) 
                min = a[i]; 
        cout << arr[min]; 
    } 
    cout << endl; 
} 
  
// 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); 
}

Python3

# Python3 program to find first   
# repeating element in arr[]  
  
# This function prints the   
# first repeating element in arr[] 
def printFirstRepeating(arr, n): 
      
    # This will set k=1, if any   
    # repeating element found  
    k = 0
  
    # max = maximum from (all elements & n)  
    max = n 
      
    for i in range(n): 
        if (max < arr[i]): 
            max = arr[i] 
      
    # Array a is for storing   
    # 1st time occurence of element  
    # initialized by 0  
    a = [0 for i in range(max + 1)] 
  
    # Store 1 in array b   
    # if element is duplicate  
    # initialized by 0 
    b = [0 for i in range(max + 1)] 
  
    for i in range(n): 
  
        # Duplicate element found  
        if (a[arr[i]]): 
            b[arr[i]] = 1
            k = 1
            continue
        else: 
  
            # Storing 1st occurence of arr[i]  
            a[arr[i]] = i 
  
    if (k == 0): 
        print("No repeating element found") 
    else: 
        min = max + 1
  
        for i in range(max + 1): 
              
            # Trace array a & find repeating  
            # element with min index  
            if (a[i] and (min > (a[i])) and b[i]): 
                min = a[i] 
                  
        print(arr[min]) 
  
# Driver code 
arr = [ 10, 5, 3, 4, 3, 5, 6 ] 
n = len(arr) 
  
printFirstRepeating(arr, n)  
  
# This code is contributed by avanitrachhadiya2155

Output

Time Complexity: O(n).

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

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C++

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