# Find the first repeating element in an array of integers

Given an array of integers arr[], The task is to find the index of first repeating element in it i.e. the element that occurs more than once and whose index of the first occurrence is the smallest.

Examples:

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

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

Recommended Practice

Naive Approach: Below is the idea to solve the problem

Run two nested loops, the outer loop picks an element one by one, and the inner loop checks whether the element is repeated or not. Once a repeating element is found, break the loops and print the element.

## C++

 // Including necessary header files #include using namespace std;   // Function to find the index of first repeating element in // an array int firstRepeatingElement(int arr[], int n) {     // Nested loop to check for repeating elements     for (int i = 0; i < n; i++) {         for (int j = i + 1; j < n; j++) {             // If a repeating element is found, return its             // index             if (arr[i] == arr[j]) {                 return i;             }         }     }     // If no repeating element is found, return -1     return -1; }   // Driver code int main() {     // Initializing an array and its size     int arr[] = { 10, 5, 3, 4, 3, 5, 6 };     int n = sizeof(arr) / sizeof(arr[0]);     // Finding the index of first repeating element     int index = firstRepeatingElement(arr, n);       // Checking if any repeating element is found or not     if (index == -1) {         cout << "No repeating element found!" << endl;     }     else {         // Printing the first repeating element and its         // index         cout << "First repeating element is " << arr[index]              << endl;     }       return 0; }

## Java

 // Java code for the approach   import java.util.*;   public class GFG {     // Function to find the index of first repeating element in an array     public static int firstRepeatingElement(int[] arr, int n) {         // Nested loop to check for repeating elements         for (int i = 0; i < n; i++) {             for (int j = i + 1; j < n; j++) {                 // If a repeating element is found, return its index                 if (arr[i] == arr[j]) {                     return i;                 }             }         }         // If no repeating element is found, return -1         return -1;     }       // Driver code     public static void main(String[] args) {         // Initializing an array and its size         int[] arr = { 10, 5, 3, 4, 3, 5, 6 };         int n = arr.length;         // Finding the index of first repeating element         int index = firstRepeatingElement(arr, n);           // Checking if any repeating element is found or not         if (index == -1) {             System.out.println("No repeating element found!");         }         else {             // Printing the first repeating element and its index             System.out.println("First repeating element is " + arr[index]);         }     } }

## Python3

 # Python3 code for the approach   # Function to find the index of first repeating element in an array def firstRepeatingElement(arr, n):   # Nested loop to check for repeating elements   for i in range(n):     for j in range(i+1, n):       # If a repeating element is found, return its index       if arr[i] == arr[j]:         return i           # If no repeating element is found, return -1   return -1   # Driver code if __name__ == '__main__':   # Initializing an array and its size   arr = [10, 5, 3, 4, 3, 5, 6]   n = len(arr)       # Finding the index of first repeating element   index = firstRepeatingElement(arr, n)       # Checking if any repeating element is found or not   if index == -1:       print("No repeating element found!")   else:       # Printing the first repeating element and its index       print("First repeating element is", arr[index])

## C#

 using System;   public class Program {     public static void Main()     {         // Initializing an array and its size         int[] arr = { 10, 5, 3, 4, 3, 5, 6 };         int n = arr.Length;           // Initializing variables to keep track of the minimum index and         // minimum value of the repeating element         int minIndex = n;         int minValue = int.MaxValue;           // Creating a dictionary to store the last seen index of each element         var dict = new System.Collections.Generic.Dictionary();           // Iterating over the array from left to right         for (int i = 0; i < n; i++)         {             int val = arr[i];               // If the element is not in the dictionary, add it with its index             if (!dict.ContainsKey(val))             {                 dict[val] = i;             }             // If the element is already in the dictionary, update the minimum index             // and minimum value if necessary             else             {                 int index = dict[val];                 if (index < minIndex || (index == minIndex && val < minValue))                 {                     minIndex = index;                     minValue = val;                 }             }         }           // Checking if any repeating element is found or not         if (minIndex == n)         {             Console.WriteLine("No repeating element found!");         }         else         {             // Printing the first repeating element and its index             Console.WriteLine("First repeating element is " + minValue + " at index " + minIndex);         }     } }

## Javascript

 function firstRepeatingElement(arr) {     // Nested loop to check for repeating elements     for (let i = 0; i < arr.length; i++) {         for (let j = i + 1; j < arr.length; j++) {             // If a repeating element is found, return its index             if (arr[i] === arr[j]) {                 return i;             }         }     }     // If no repeating element is found, return -1     return -1; }   // Driver code const arr = [10, 5, 3, 4, 3, 5, 6]; // Finding the index of first repeating element const index = firstRepeatingElement(arr);   // Checking if any repeating element is found or not if (index === -1) {     console.log("No repeating element found!"); } else {     // Printing the first repeating element and its index     console.log("First repeating element is", arr[index]); }

Output

First repeating element is 5

Time Complexity: O(N2)
Auxiliary Space: O(1)

## Find the first repeating element in an array of integers using sorting:

Below is the idea to solve the problem.

Store the elements of arr[] in a duplicate array temp[], sort temp[] and traverse arr[] from 0 to N – 1, Simultaneously check the count of this element in temp[] and if the current element arr[i] has more than one occurrence then return arr[i].

Follow the steps below to Implement the idea:

• Copy the given array to an auxiliary array temp[] and sort temp array.
• Traverse the input array arr[] from 0 to N – 1
• If no repeating element is found print “No Repeating Number Found”.

Time complexity: O(NlogN).
Auxiliary Space: O(N)

## Find the first repeating element in an array of integers usingHashset

Below is the idea to solve the problem

The idea is to traverse the given array arr[] from right to left and update the minimum index whenever, an already visited element has been found. To check if the element was already visited Hashset can be used.

Follow the steps below to implement the idea:

• Initialize an empty Hashset myset and a variable min with -1.
• Run a for loop for each index of array arr[] from N – 1 to 0.
• If the current element is present in myset then update min with i.
• Else insert arr[i] in myset.
• Return min.

Below is the implementation of the above approach.

## C++

 /* C++ program to find first repeating element in arr[] */ #include 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 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 Code 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 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 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) {             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

## Javascript



Output

The first repeating element is 5

Time Complexity: O(n).
Auxiliary Space: O(n).

Thanks to Mohammad Shahid for suggesting this solution.

## Find the first repeating element in an array of integers usingHashing

The idea is to use Hash array to store the occurrence of elements. Then traverse the array from left to right and return the first element with occurrence more than 1.

Follow the below steps to implement the idea:

• Initialize variables k with 0, max with -1 and min with max + 1 and iterate over all values of arr[] to store the largest value in max.
• Initialize a Hash arrays a[] and b[] of size max + 1.
• Run a for loop from 0 to N – 1
• If a[arr[i]] is 0 put i+1 in place of a[arr[i]].
• Else assign 1 to b[arrr[i]] and k.
• If k is 0 print “No repeating element found”.
• Else iterate from 0 to max
• If a[i] is not zero and b[i] is not zero and min is greater than a[i] then update min a[i].
• Print min.

Below is the Implementation of above approach

## C++

 /* C++ program to find first repeating element in arr[] */ #include 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 occurrence 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 occurrence of arr[i]             a[arr[i]] = i + 1;     }       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 - 1];     }     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); }

## Java

 /* Java program to find first repeating element in arr[] */ public class GFG {       // This function prints the     // first repeating element in arr[]     static 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 occurrence of element         // initialized by 0         int[] a = new int[max + 1];           // Store 1 in array b         // if element is duplicate         // initialized by 0         int[] b = new int[max + 1];         for (int i = 0; i < n; i++) {               // Duplicate element found             if (a[arr[i]] != 0) {                 b[arr[i]] = 1;                 k = 1;                 continue;             }             else                 // storing 1st occurrence of arr[i]                 a[arr[i]] = i + 1;         }           if (k == 0)             System.out.println(                 "No repeating element found");         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] != 0 && min > a[i] && b[i] != 0)                     min = a[i];             System.out.print(arr[min - 1]);         }         System.out.println();     }       // Driver code     public static void main(String[] args)     {         int[] arr = { 10, 5, 3, 4, 3, 5, 6 };           int N = arr.length;         printFirstRepeating(arr, N);     } }   // This code is contributed by divyesh072019

## 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 occurrence 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 occurrence of arr[i]             a[arr[i]] = i+1       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-1])     # Driver code arr = [10, 5, 3, 4, 3, 5, 6] N = len(arr)   printFirstRepeating(arr, N)   # This code is contributed by avanitrachhadiya2155

## C#

 /* C# program to find first repeating element in arr[] */ using System; class GFG {       // This function prints the     // first repeating element in arr[]     static 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 occurrence of element         // initialized by 0         int[] a = new int[max + 1];           // Store 1 in array b         // if element is duplicate         // initialized by 0         int[] b = new int[max + 1];           for (int i = 0; i < n; i++) {               // Duplicate element found             if (a[arr[i]] != 0) {                 b[arr[i]] = 1;                 k = 1;                 continue;             }             else                 // storing 1st occurrence of arr[i]                 a[arr[i]] = i + 1;         }           if (k == 0)             Console.WriteLine("No repeating element found");         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] != 0) && min > a[i]                     && (b[i] != 0))                     min = a[i];             Console.Write(arr[min - 1]);         }         Console.WriteLine();     }       // Driver code     static void Main()     {         int[] arr = { 10, 5, 3, 4, 3, 5, 6 };           int N = arr.Length;         printFirstRepeating(arr, N);     } }   // This code is contributed by divyeshrabadiya07.

## Javascript



Output

5

Time Complexity: O(N).
Auxiliary Space: O(N).

#### Another approach using single hash array

Follow the below steps to implement the idea:

• Initialize a variable max to -1 to keep track of the maximum value in the array.
• Iterate over all values of arr[] to store the largest value in max.
• Declare an integer array hash of size max+1 and initialize all its elements to 0. This array will be used as a hash table to store the count of occurrences of each element in the input array.
• Traverse the input array again from index 0 to n-1, and increment the count of the corresponding element in the hash table.
• Traverse the input array again from index 0 to n-1, and for each element in the input array, check if the count of the corresponding element in the hash table is greater than 1. If it is, return the index of that element in the input array (i.e., i+1, since the function is expected to return a 1-based index). If no repeated element is found, the function returns -1.

Below is the Implementation of above approach

## C++

 /* C++ program to find first repeating element in arr[] */ #include using namespace std;   // This function prints the // first repeating element in arr[] void firstRepeating(int arr[], int n) {               int max = -1;           //Finding max         for(int i = 0 ; i1){                 repeating=arr[i];                   break;             }         }           if(repeating==INT_MIN){           cout << "There are no repeating elements";         }           else{           cout << "The first repeating element is : "              <

## Java

 /*package whatever //do not write package name here */   import java.io.*;   class GFG {       static void firstRepeating(int[] arr, int n) {         int max = -1;         // Finding max         for (int i = 0; i < n; i++) {             if (max < arr[i]) {                 max = arr[i];             }         }           // Creating array         int[] hash = new int[max + 1];           // Mapping/counting         for (int i = 0; i < n; i++) {             hash[arr[i]]++;         }           // Checking for repeating element         int repeating = Integer.MIN_VALUE;         for (int i = 0; i < n; i++) {             if (hash[arr[i]] > 1) {                 repeating = arr[i];                 break;             }         }           if (repeating == Integer.MIN_VALUE) {             System.out.println("There are no repeating elements");         } else {             System.out.println("The first repeating element is : " + repeating);         }     }     public static void main (String[] args) {         int[] arr = { 10, 5, 3, 4, 3, 5, 6 };         int N = arr.length;         firstRepeating(arr, N);     } }

## Python3

 # Python program to find first repeating element in arr[] def firstRepeating(arr, n):     # Finding max     max_val = max(arr)           # Creating array     hash = [0] * (max_val+1)           # Mapping/counting     for i in range(n):         hash[arr[i]] += 1           # Variable for storing ans     repeating = float('inf')           # Checking repeating element     for i in range(n):         if hash[arr[i]] > 1:             repeating = arr[i]             break           if repeating == float('inf'):         print("There are no repeating elements")     else:         print("The first repeating element is:", repeating)   # Driver method to test above method arr = [10, 5, 3, 4, 3, 5, 6] N = len(arr) firstRepeating(arr, N)

## C#

 // C# program to find first // repeating element in arr[] using System;   public class Program {     // This function prints the   // first repeating element in arr[]   public static void FirstRepeating(int[] arr, int n)   {     int max = -1;           // Finding max     for (int i = 0; i < n; i++) {       if (max < arr[i]) {         max = arr[i];       }     }       // Creating array     int[] hash = new int[max + 1];       // Mapping/counting     for (int i = 0; i < n; i++) {       hash[arr[i]]++;     }       // Variable for storing ans     int repeating = int.MinValue;       // Checking repeating element     for (int i = 0; i < n; i++) {       if (hash[arr[i]] > 1) {         repeating = arr[i];         break;       }     }       if (repeating == int.MinValue) {       Console.WriteLine(         "There are no repeating elements");     }     else {       Console.WriteLine(         "The first repeating element is: "         + repeating);     }   }     // Driver method to test above method   public static void Main()   {     int[] arr = { 10, 5, 3, 4, 3, 5, 6 };     int n = arr.Length;     FirstRepeating(arr, n);   } }   // This code is contributed by Susobhan Akhuli

## Javascript

 function firstRepeating(arr, n) {     // Finding max     let max_val = Math.max(...arr);       // Creating array     let hash = new Array(max_val + 1).fill(0);       // Mapping/counting     for (let i = 0; i < n; i++) {         hash[arr[i]] += 1;     }       // Variable for storing ans     let repeating = Infinity;       // Checking repeating element     for (let i = 0; i < n; i++) {         if (hash[arr[i]] > 1) {             repeating = arr[i];             break;         }     }       if (repeating == Infinity) {         console.log("There are no repeating elements");     } else {         console.log("The first repeating element is:", repeating);     } }   // Driver method to test above method let arr = [10, 5, 3, 4, 3, 5, 6]; let N = arr.length; firstRepeating(arr,N);

Output

The first repeating element is : 5

Time Complexity: O(N).
Auxiliary Space: O(N).

The first for loop that finds the maximum element in the array has a time complexity of O(n). The second for loop that creates a hash array has a time complexity of O(n). The third for loop that checks for the first repeating element also has a time complexity of O(n). The array named ‘hash’ is created with max+1 elements so space O(max+1).

Since all three loops run sequentially, the total time complexity of the code is O(n).

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