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 : 2Input: {9, 4, 9, 6, 7, 4}
Output: 6
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 <bits/stdc++.h> 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
<script> // JavaScript code for the above approach function firstNonRepeating(arr, n) { // Loop for checking each element for (let i = 0; i < n; i++) { let 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 let arr = [9, 4, 9, 6, 7, 4]; let n = arr.length; document.write(firstNonRepeating(arr, n)); // This code is contributed by Potta Lokesh </script> |
PHP
<?php // Simple PHP program to find first non- // repeating element. function firstNonRepeating( $arr , $n ) { // Loop for checking each element for ( $i = 0; $i < $n ; $i ++) { $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 $arr = array (9, 4, 9, 6, 7, 4); $n = sizeof( $arr ) ; echo firstNonRepeating( $arr , $n ); // This code is contributed by ajit ?> |
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 <bits/stdc++.h> 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<Integer, Integer> 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
<script> // Efficient javascript program to find first non- // repeating element. function firstNonRepeating(arr , n) { // Insert all array elements in hash // table const m = new Map(); for ( var i = 0; i < n; i++) { if (m.has(arr[i])) { m.set(arr[i], m.get(arr[i]) + 1); } else { m.set(arr[i], 1); } } // Traverse array again and return // first element with count 1. for ( var i = 0; i < n; i++) if (m.get(arr[i]) == 1) return arr[i]; return -1; } // Driver code var arr = [ 9, 4, 9, 6, 7, 4 ]; var n = arr.length; document.write(firstNonRepeating(arr, n)); // This code contributed by shikhasingrajput </script> |
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