Find the first repeating element in an array of integers
Given an array of integers, find the first repeating element in it. We need to find the element that occurs more than once and whose index of first occurrence is smallest.
Examples:
Input: arr[] = {10, 5, 3, 4, 3, 5, 6}
Output: 5 [5 is the first element that repeats]
Input: arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10}
Output: 6 [6 is the first element that repeats]
A Simple Solution is to use two nested loops. The outer loop picks an element one by one, the inner loop checks whether the element is repeated or not. Once we find an element that repeats, we break the loops and print the element. Time Complexity of this solution is O(n2)
We can Use Sorting to solve the problem in O(nLogn) time. Following are detailed steps.
1) Copy the given array to an auxiliary array temp[].
2) Sort the temp array using a O(nLogn) time sorting algorithm.
3) Scan the input array from left to right. For every element, count its occurrences in temp[] using binary search. As soon as we find an element that occurs more than once, we return the element. This step can be done in O(nLogn) time.
We can Use Hashing to solve this in O(n) time on average. The idea is to traverse the given array from right to left and update the minimum index whenever we find an element that has been visited on right side. Thanks to Mohammad Shahid for suggesting this solution.
Following is the implementation of this idea.
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 method to test above methodint 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 Codearr = [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 |
Output
The first repeating element is 5
Another Approach:
If you want to do this without using any additional data structure. The problem can also be solved using array only. See the method below.
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 occurence 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 occurence of arr[i] a[arr[i]] = i; } 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]; } cout << endl;}// Driver method to test above methodint 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 occurence 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 occurence of arr[i] a[arr[i]] = i; } 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]); } 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 occurence 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 occurence of arr[i] a[arr[i]] = i 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])# Driver codearr = [ 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 occurence 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 occurence of arr[i] a[arr[i]] = i; } 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]); } 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. |
Output
5
Time Complexity: O(n).
Another Approach By Using O(n) Auxilliary Space and O(n) Time Complexity:
Java
/*package whatever //do not write package name here */import java.io.*;class GFG { public static int firstRepeated(int[] arr, int n) { int max = 0; for (int x = 0; x < n; x++) { if (arr[x] > max) { max = arr[x]; } } int temp[] = new int[max + 1]; // the idea is to use // temporary array as hashmap Arrays.fill(temp, 0); for (int x = 0; x < n; x++) { int num = arr[x]; temp[num]++; } for (int x = 0; x < n; x++) { int num = arr[x]; if (temp[num] > 1) { return x; } } return -1; // if no repeating element found } public static void main(String[] args) { int[] arr = { 10, 5, 3, 4, 3, 5, 6 }; int n = arr.length; int index = firstRepeated( arr, arr.length); // index Of 1st repeating number if (index != -1) { System.out.println("1st Repeatig Number is " + arr[index]); } else { System.out.println("No Repeating Number Found"); } }} |
Another Approach Using Python inbuilt functions:
Python3
# Python3 program to find first# repeating element in arr[]# This function prints the# first repeating element in arr[]def printFirstRepeating(a, n): for i in range(len(a)): if a.count(a[i]) > 1: return a[i] return "there is no repetition number"# Driver codearr = [10, 5, 3, 4, 3, 5, 6]n = len(arr)print(printFirstRepeating(arr, n))# This code is contributed by karthikeyakumarnallam |
Output
5
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