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: 5
Explanation: 5 is the first element that repeatsInput: arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10}
Output: 6
Explanation: 6 is the first element that repeats
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.
- For every element, count its occurrences in temp[] using binary search.
- If the count of occurrence of current element is more than one, then return the current element.
- 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 using Hashset
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 <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 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<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 |
Javascript
<script> // Javascript program to find first // repeating element in arr[] // This function prints the first // repeating element in arr[] function printFirstRepeating(arr) { // Initialize index of first // repeating element let min = -1; // Creates an empty hashset let set = new Set(); // Traverse the input array from right to left for (let i = arr.length - 1; i >= 0; i--) { // If element is already in // hash set, update min if (set.has(arr[i])) min = i; // Else add element to hash set else set.add(arr[i]); } // Print the result if (min != -1) document.write( "The first repeating element is " + arr[min]); else document.write( "There are no repeating elements" ); } // Driver code let arr = [ 10, 5, 3, 4, 3, 5, 6 ]; printFirstRepeating(arr); // This code is contributed by unknown2108 </script> |
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 using Hashing
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 <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 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
<script> /* javascript program to find first repeating element in arr */ // This function prints the // first repeating element in arr function printFirstRepeating(arr , n) { // This will set k=1, if any // repeating element found var k = 0; // max = maximum from (all elements & n) var max = n; for (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 var a = Array(max + 1).fill(0); // Store 1 in array b // if element is duplicate // initialized by 0 var b = Array(max + 1).fill(0); for ( var 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) document.write( "No repeating element found" ); else { var min = max + 1; // trace array a & find repeating element // with min index for (i = 0; i < max + 1; i++) if (a[i] != 0 && min > a[i] && b[i] != 0) min = a[i]; document.write(arr[min-1]); } document.write( "<br/>" ); } // Driver code var arr = [ 10, 5, 3, 4, 3, 5, 6 ]; var N = arr.length; printFirstRepeating(arr, N); // This code is contributed by todaysgaurav </script> |
5
Time Complexity: O(N).
Auxiliary Space: O(N).
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