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# Find the first repeating element in an array of integers

• Difficulty Level : Easy
• Last Updated : 14 Dec, 2022

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.

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<``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 Code``int` `main()``{``    ``int` `arr[] = { 10, 5, 3, 4, 3, 5, 6 };`` ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``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<``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`

## 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);``    ``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).

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