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.
C++
// Including necessary header files#include <bits/stdc++.h>using namespace std;// Function to find the index of first repeating element in// an arrayint firstRepeatingElement(int arr[], int n){ // Nested loop to check for repeating elements for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // If a repeating element is found, return its // index if (arr[i] == arr[j]) { return i; } } } // If no repeating element is found, return -1 return -1;}// Driver codeint main(){ // Initializing an array and its size int arr[] = { 10, 5, 3, 4, 3, 5, 6 }; int n = sizeof(arr) / sizeof(arr[0]); // Finding the index of first repeating element int index = firstRepeatingElement(arr, n); // Checking if any repeating element is found or not if (index == -1) { cout << "No repeating element found!" << endl; } else { // Printing the first repeating element and its // index cout << "First repeating element is " << arr[index] << endl; } return 0;} |
Java
// Java code for the approachimport java.util.*;public class GFG { // Function to find the index of first repeating element in an array public static int firstRepeatingElement(int[] arr, int n) { // Nested loop to check for repeating elements for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // If a repeating element is found, return its index if (arr[i] == arr[j]) { return i; } } } // If no repeating element is found, return -1 return -1; } // Driver code public static void main(String[] args) { // Initializing an array and its size int[] arr = { 10, 5, 3, 4, 3, 5, 6 }; int n = arr.length; // Finding the index of first repeating element int index = firstRepeatingElement(arr, n); // Checking if any repeating element is found or not if (index == -1) { System.out.println("No repeating element found!"); } else { // Printing the first repeating element and its index System.out.println("First repeating element is " + arr[index]); } }} |
Python3
# Python3 code for the approach# Function to find the index of first repeating element in an arraydef firstRepeatingElement(arr, n): # Nested loop to check for repeating elements for i in range(n): for j in range(i+1, n): # If a repeating element is found, return its index if arr[i] == arr[j]: return i # If no repeating element is found, return -1 return -1# Driver codeif __name__ == '__main__': # Initializing an array and its size arr = [10, 5, 3, 4, 3, 5, 6] n = len(arr) # Finding the index of first repeating element index = firstRepeatingElement(arr, n) # Checking if any repeating element is found or not if index == -1: print("No repeating element found!") else: # Printing the first repeating element and its index print("First repeating element is", arr[index]) |
C#
using System;public class Program{ public static void Main() { // Initializing an array and its size int[] arr = { 10, 5, 3, 4, 3, 5, 6 }; int n = arr.Length; // Initializing variables to keep track of the minimum index and // minimum value of the repeating element int minIndex = n; int minValue = int.MaxValue; // Creating a dictionary to store the last seen index of each element var dict = new System.Collections.Generic.Dictionary<int, int>(); // Iterating over the array from left to right for (int i = 0; i < n; i++) { int val = arr[i]; // If the element is not in the dictionary, add it with its index if (!dict.ContainsKey(val)) { dict[val] = i; } // If the element is already in the dictionary, update the minimum index // and minimum value if necessary else { int index = dict[val]; if (index < minIndex || (index == minIndex && val < minValue)) { minIndex = index; minValue = val; } } } // Checking if any repeating element is found or not if (minIndex == n) { Console.WriteLine("No repeating element found!"); } else { // Printing the first repeating element and its index Console.WriteLine("First repeating element is " + minValue + " at index " + minIndex); } }} |
Javascript
function firstRepeatingElement(arr) { // Nested loop to check for repeating elements for (let i = 0; i < arr.length; i++) { for (let j = i + 1; j < arr.length; j++) { // If a repeating element is found, return its index if (arr[i] === arr[j]) { return i; } } } // If no repeating element is found, return -1 return -1;}// Driver codeconst arr = [10, 5, 3, 4, 3, 5, 6];// Finding the index of first repeating elementconst index = firstRepeatingElement(arr);// Checking if any repeating element is found or notif (index === -1) { console.log("No repeating element found!");} else { // Printing the first repeating element and its index console.log("First repeating element is", arr[index]);} |
Output
First repeating element is 5
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 Codeint 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 |
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 codelet arr = [ 10, 5, 3, 4, 3, 5, 6 ];printFirstRepeating(arr);// This code is contributed by unknown2108 </script> |
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 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 firstrepeating 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 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 firstrepeating 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 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 firstrepeating 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> |
Output
5
Time Complexity: O(N).
Auxiliary Space: O(N).
Another approach using single hash array
Follow the below steps to implement the idea:
- Initialize a variable max to -1 to keep track of the maximum value in the array.
- Iterate over all values of arr[] to store the largest value in max.
- Declare an integer array hash of size max+1 and initialize all its elements to 0. This array will be used as a hash table to store the count of occurrences of each element in the input array.
- Traverse the input array again from index 0 to n-1, and increment the count of the corresponding element in the hash table.
- Traverse the input array again from index 0 to n-1, and for each element in the input array, check if the count of the corresponding element in the hash table is greater than 1. If it is, return the index of that element in the input array (i.e., i+1, since the function is expected to return a 1-based index). If no repeated element is found, the function returns -1.
Below is the Implementation of above approach
C++
/* C++ program to find firstrepeating element in arr[] */#include <bits/stdc++.h>using namespace std;// This function prints the// first repeating element in arr[]void firstRepeating(int arr[], int n) { int max = -1; //Finding max for(int i = 0 ; i<n;i++){ if(max<arr[i]){ max = arr[i]; } } //Creating array int hash[max+1]={0}; //Mapping/counting for(int i =0;i<n; i++){ hash[arr[i]]++; } //Variable for storing ans int repeating=INT_MIN; //Checking repeatibng element for(int i =0;i<n; i++){ if(hash[arr[i]]>1){ repeating=arr[i]; break; } } if(repeating==INT_MIN){ cout << "There are no repeating elements"; } else{ cout << "The first repeating element is : " <<repeating; } }// Driver method to test above methodint main(){ int arr[] = { 10, 5, 3, 4, 3, 5, 6 }; int N = sizeof(arr) / sizeof(arr[0]); firstRepeating(arr, N);} |
Java
/*package whatever //do not write package name here */import java.io.*;class GFG { static void firstRepeating(int[] arr, int n) { int max = -1; // Finding max for (int i = 0; i < n; i++) { if (max < arr[i]) { max = arr[i]; } } // Creating array int[] hash = new int[max + 1]; // Mapping/counting for (int i = 0; i < n; i++) { hash[arr[i]]++; } // Checking for repeating element int repeating = Integer.MIN_VALUE; for (int i = 0; i < n; i++) { if (hash[arr[i]] > 1) { repeating = arr[i]; break; } } if (repeating == Integer.MIN_VALUE) { System.out.println("There are no repeating elements"); } else { System.out.println("The first repeating element is : " + repeating); } } public static void main (String[] args) { int[] arr = { 10, 5, 3, 4, 3, 5, 6 }; int N = arr.length; firstRepeating(arr, N); }} |
Python3
# Python program to find first repeating element in arr[]def firstRepeating(arr, n): # Finding max max_val = max(arr) # Creating array hash = [0] * (max_val+1) # Mapping/counting for i in range(n): hash[arr[i]] += 1 # Variable for storing ans repeating = float('inf') # Checking repeating element for i in range(n): if hash[arr[i]] > 1: repeating = arr[i] break if repeating == float('inf'): print("There are no repeating elements") else: print("The first repeating element is:", repeating)# Driver method to test above methodarr = [10, 5, 3, 4, 3, 5, 6]N = len(arr)firstRepeating(arr, N) |
C#
// C# program to find first// repeating element in arr[]using System;public class Program { // This function prints the // first repeating element in arr[] public static void FirstRepeating(int[] arr, int n) { int max = -1; // Finding max for (int i = 0; i < n; i++) { if (max < arr[i]) { max = arr[i]; } } // Creating array int[] hash = new int[max + 1]; // Mapping/counting for (int i = 0; i < n; i++) { hash[arr[i]]++; } // Variable for storing ans int repeating = int.MinValue; // Checking repeating element for (int i = 0; i < n; i++) { if (hash[arr[i]] > 1) { repeating = arr[i]; break; } } if (repeating == int.MinValue) { Console.WriteLine( "There are no repeating elements"); } else { Console.WriteLine( "The first repeating element is: " + repeating); } } // Driver method to test above method public static void Main() { int[] arr = { 10, 5, 3, 4, 3, 5, 6 }; int n = arr.Length; FirstRepeating(arr, n); }}// This code is contributed by Susobhan Akhuli |
Javascript
function firstRepeating(arr, n) { // Finding max let max_val = Math.max(...arr); // Creating array let hash = new Array(max_val + 1).fill(0); // Mapping/counting for (let i = 0; i < n; i++) { hash[arr[i]] += 1; } // Variable for storing ans let repeating = Infinity; // Checking repeating element for (let i = 0; i < n; i++) { if (hash[arr[i]] > 1) { repeating = arr[i]; break; } } if (repeating == Infinity) { console.log("There are no repeating elements"); } else { console.log("The first repeating element is:", repeating); }}// Driver method to test above methodlet arr = [10, 5, 3, 4, 3, 5, 6];let N = arr.length;firstRepeating(arr,N); |
Output
The first repeating element is : 5
Time Complexity: O(N).
Auxiliary Space: O(N).
The first for loop that finds the maximum element in the array has a time complexity of O(n). The second for loop that creates a hash array has a time complexity of O(n). The third for loop that checks for the first repeating element also has a time complexity of O(n). The array named ‘hash’ is created with max+1 elements so space O(max+1).
Since all three loops run sequentially, the total time complexity of the code is O(n).
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