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# Find first non-repeating element in a given Array of integers

Given an array of integers of size N, the task is to find the first non-repeating element in this array.

Examples:

Input: {-1, 2, -1, 3, 0}
Output: 2
Explanation: The first number that does not repeat is : 2

Input: {9, 4, 9, 6, 7, 4}
Output: 6

Recommended Practice

## Find first non-repeating element in a given Array of integers using Nested Loops:

This approach is based on the following idea:

Simple Solution is to use two loops. The outer loop picks elements one by one and the inner loop checks if the element is present more than once or not..

Illustration:

Given arr[] = {-1, 2, -1, 3, 0}

For element at i = 0

• The value of element at index 2 is same, then this can’t be first non-repeating element

For element at i = 1:

• After traversing the array arr[1] is not present is not present in the array except at 1.

Hence, element is index 1 is the first non-repeating element which is 2

Follow the steps below to solve the given problem:

• Loop over the array from the left.
• Check for each element if its presence is present in the array for more than 1 time.
• Use a nested loop to check the presence.

Below is the implementation of the above idea:

## C++

 `// Simple CPP program to find first non-``// repeating element.``#include ``using` `namespace` `std;` `int` `firstNonRepeating(``int` `arr[], ``int` `n)``{``    ``// Loop for checking each element``    ``for` `(``int` `i = 0; i < n; i++) {``        ``int` `j;``        ``// Checking if ith element is present in array``        ``for` `(j = 0; j < n; j++)``            ``if` `(i != j && arr[i] == arr[j])``                ``break``;``        ``// if ith element is not present in array``        ``// except at ith index then return element``        ``if` `(j == n)``            ``return` `arr[i];``    ``}``    ``return` `-1;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 9, 4, 9, 6, 7, 4 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << firstNonRepeating(arr, n);``    ``return` `0;``}`

## Java

 `// Java program to find first non-repeating``// element.``class` `GFG {` `    ``static` `int` `firstNonRepeating(``int` `arr[], ``int` `n)``    ``{``        ``// Loop for checking each element``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``int` `j;``            ``// Checking if ith element is present in array``            ``for` `(j = ``0``; j < n; j++)``                ``if` `(i != j && arr[i] == arr[j])``                    ``break``;``            ``// if ith element is not present in array``            ``// except at ith index then return element``            ``if` `(j == n)``                ``return` `arr[i];``        ``}` `        ``return` `-``1``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{` `        ``int` `arr[] = { ``9``, ``4``, ``9``, ``6``, ``7``, ``4` `};``        ``int` `n = arr.length;` `        ``System.out.print(firstNonRepeating(arr, n));``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python3 program to find first``# non-repeating element.`  `def` `firstNonRepeating(arr, n):` `    ``# Loop for checking each element``    ``for` `i ``in` `range``(n):``        ``j ``=` `0``        ``# Checking if ith element is present in array``        ``while``(j < n):``            ``if` `(i !``=` `j ``and` `arr[i] ``=``=` `arr[j]):``                ``break``            ``j ``+``=` `1``        ``# if ith element is not present in array``        ``# except at ith index then return element``        ``if` `(j ``=``=` `n):``            ``return` `arr[i]` `    ``return` `-``1`  `# Driver code``arr ``=` `[``9``, ``4``, ``9``, ``6``, ``7``, ``4``]``n ``=` `len``(arr)``print``(firstNonRepeating(arr, n))` `# This code is contributed by Anant Agarwal.`

## C#

 `// C# program to find first non-``// repeating element.``using` `System;` `class` `GFG {``    ``static` `int` `firstNonRepeating(``int``[] arr, ``int` `n)``    ``{``        ``// Loop for checking each element``        ``for` `(``int` `i = 0; i < n; i++) {``            ``int` `j;``            ``// Checking if ith element is present in array``            ``for` `(j = 0; j < n; j++)``                ``if` `(i != j && arr[i] == arr[j])``                    ``break``;``            ``// if ith element is not present in array``            ``// except at ith index then return element``            ``if` `(j == n)``                ``return` `arr[i];``        ``}``        ``return` `-1;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr = { 9, 4, 9, 6, 7, 4 };``        ``int` `n = arr.Length;``        ``Console.Write(firstNonRepeating(arr, n));``    ``}``}``// This code is contributed by Anant Agarwal.`

## JavaScript

 ``

## PHP

 ``

Output

```6

```

Time Complexity: O(n*n), Checking for each element n times
Auxiliary Space: O(1)

## Find first non-repeating element in a given Array of integers using Hashing:

This approach is based on the following idea:

• The idea is to store the frequency of every element in the hashmap.
• Then check the first element whose frequency is 1 in the hashmap.
• This can be achieved using hashing

Illustration:

arr[] = {-1, 2, -1, 3, 0}

Frequency map for arr:

• -1 -> 2
• 2 -> 1
• 3 -> 1
• 0 -> 1

Traverse arr[] from left:

At i = 0:

• Frequency of arr[0] is 2, therefore it can’t be first non-repeating element

At i = 1:

• Frequency of arr[1] is 1, therefore it will be the first non-repeating element.

Hence, 2 is the first non-repeating element.

Follow the steps below to solve the given problem:

• Traverse array and insert elements and their counts in the hash table.
• Traverse array again and print the first element with a count equal to 1.

Below is the implementation of the above idea:

## C++

 `// Efficient CPP program to find first non-``// repeating element.``#include ``using` `namespace` `std;` `int` `firstNonRepeating(``int` `arr[], ``int` `n)``{``    ``// Insert all array elements in hash``    ``// table``    ``unordered_map<``int``, ``int``> mp;``    ``for` `(``int` `i = 0; i < n; i++)``        ``mp[arr[i]]++;` `    ``// Traverse array again and return``    ``// first element with count 1.``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `(mp[arr[i]] == 1)``            ``return` `arr[i];``    ``return` `-1;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 9, 4, 9, 6, 7, 4 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << firstNonRepeating(arr, n);``    ``return` `0;``}`

## Java

 `// Efficient Java program to find first non-``// repeating element.``import` `java.util.*;` `class` `GFG {` `    ``static` `int` `firstNonRepeating(``int` `arr[], ``int` `n)``    ``{``        ``// Insert all array elements in hash``        ``// table` `        ``Map m = ``new` `HashMap<>();``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``if` `(m.containsKey(arr[i])) {``                ``m.put(arr[i], m.get(arr[i]) + ``1``);``            ``}``            ``else` `{``                ``m.put(arr[i], ``1``);``            ``}``        ``}``        ``// Traverse array again and return``        ``// first element with count 1.``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``if` `(m.get(arr[i]) == ``1``)``                ``return` `arr[i];``        ``return` `-``1``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``9``, ``4``, ``9``, ``6``, ``7``, ``4` `};``        ``int` `n = arr.length;``        ``System.out.println(firstNonRepeating(arr, n));``    ``}``}` `// This code contributed by Rajput-Ji`

## Python3

 `# Efficient Python3 program to find first``# non-repeating element.``from` `collections ``import` `defaultdict`  `def` `firstNonRepeating(arr, n):``    ``mp ``=` `defaultdict(``lambda``: ``0``)` `    ``# Insert all array elements in hash table``    ``for` `i ``in` `range``(n):``        ``mp[arr[i]] ``+``=` `1` `    ``# Traverse array again and return``    ``# first element with count 1.``    ``for` `i ``in` `range``(n):``        ``if` `mp[arr[i]] ``=``=` `1``:``            ``return` `arr[i]``    ``return` `-``1`  `# Driver Code``arr ``=` `[``9``, ``4``, ``9``, ``6``, ``7``, ``4``]``n ``=` `len``(arr)``print``(firstNonRepeating(arr, n))` `# This code is contributed by Shrikant13`

## C#

 `// Efficient C# program to find first non-``// repeating element.``using` `System;``using` `System.Collections.Generic;` `class` `GFG {` `    ``static` `int` `firstNonRepeating(``int``[] arr, ``int` `n)``    ``{``        ``// Insert all array elements in hash``        ``// table` `        ``Dictionary<``int``, ``int``> m = ``new` `Dictionary<``int``, ``int``>();``        ``for` `(``int` `i = 0; i < n; i++) {``            ``if` `(m.ContainsKey(arr[i])) {``                ``var` `val = m[arr[i]];``                ``m.Remove(arr[i]);``                ``m.Add(arr[i], val + 1);``            ``}``            ``else` `{``                ``m.Add(arr[i], 1);``            ``}``        ``}` `        ``// Traverse array again and return``        ``// first element with count 1.``        ``for` `(``int` `i = 0; i < n; i++)``            ``if` `(m[arr[i]] == 1)``                ``return` `arr[i];``        ``return` `-1;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] arr = { 9, 4, 9, 6, 7, 4 };``        ``int` `n = arr.Length;``        ``Console.WriteLine(firstNonRepeating(arr, n));``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Javascript

 ``

Output

```6

```

Time Complexity: O(n), Traverse over the array to map the frequency and again traverse over the array to check for frequency.
Auxiliary Space: O(n), Create a hash table for storing frequency