Find the first repeating element in an array of integers
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18 Nov, 2024
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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 – O(n^2) Time and O(1) Space
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.
// Cpp to find first repeating element
// using Naive approach in O(n^2) Time and O(1) Space
#include <bits/stdc++.h>
using namespace std;
// Function to find the index of the first
// repeating element in a vector
int firstRepeatingElement(vector<int> &arr)
{
int n = arr.size();
// Nested loop to check for repeating elements
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (arr[i] == arr[j]) {
return i;
}
}
}
// If no repeating element is found, return -1
return -1;
}
int main()
{
vector<int> arr = {10, 5, 3, 4, 3, 5, 6};
int index = firstRepeatingElement(arr);
if (index == -1)
cout << "No repeating found!" << endl;
else
cout << "First repeating is " << arr[index] << endl;
return 0;
}
// Java program to find first repeating element
// using Naive approach in O(n^2) Time and O(1) Space
import java.util.*;
class GfG {
// Function to find the index of first repeating element in an array
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 {
System.out.println("First repeating element is " + arr[index]);
}
}
}
# Python program to find first repeating element
# using Naive approach in O(n^2) Time and O(1) Space
# Function to find the index of first repeating element in an array
def 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 code
if __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:
print("First repeating element is", arr[index])
// C# program to find first repeating element
// using Naive approach in O(n^2) Time and O(1) Space
using System;
class GfG
{
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
{
Console.WriteLine("First repeating element is " + minValue + " at index " + minIndex);
}
}
}
// JavaScript program to find first repeating element
// using Naive approach in O(n^2) Time and O(1) Space
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;
}
const arr = [10, 5, 3, 4, 3, 5, 6];
// Finding the index of first repeating element
const index = firstRepeatingElement(arr);
if (index === -1) {
console.log("No repeating element found!");
} else {
console.log("First repeating element is", arr[index]);
}
Output
First repeating is 5
Expected Approach – Hash Set – O(n) Time and O(n) Space
Follow the steps below to implement the idea:
- Initialize an empty Hash Set s
- Run a for loop for each index i
- If the current element is present in the hash set, then return i
- Else insert arr[i] in s.
- Return -1 if no repeating element found
Below is the implementation of the above approach.
// Cpp program to find first repeating element
// using Hash Set in O(n) Time and O(n) Space
#include <iostream>
#include <vector>
#include <unordered_set>
#include <climits>
using namespace std;
int firstRepeatingElement(const vector<int>& arr) {
unordered_set<int> s;
// If an element is already present, return it
// else insert it
int minEle = INT_MAX;
for (int i = arr.size() - 1; i >= 0; i--) {
if (s.find(arr[i]) != s.end()) {
minEle = min(minEle, i);
}
s.insert(arr[i]);
}
// If no element repeats
return minEle == INT_MAX? -1: minEle;
}
int main()
{
vector<int> arr = {10, 5, 3, 4, 3, 5, 6};
int index = firstRepeatingElement(arr);
if (index == -1)
cout << "No repeating found!" << endl;
else
cout << "First repeating is " << arr[index] << endl;
return 0;
}
// Java program to find first repeating element
// using Hash Set in O(n) Time and O(n) Space
import java.util.*;
class GfG {
static int firstRepeatingElement(int[] arr) {
HashSet<Integer> s = new HashSet<>();
// If an element is already present, return it
// else insert it
int minEle = Integer.MAX_VALUE;
for (int i = arr.length - 1; i >= 0; i--) {
if (s.contains(arr[i])) {
minEle = Math.min(minEle, i);
}
s.add(arr[i]);
}
// If no element repeats
return minEle == Integer.MAX_VALUE ? -1 : minEle;
}
public static void main(String[] args) {
int[] arr = {10, 5, 3, 4, 3, 5, 6};
int index = firstRepeatingElement(arr);
if (index == -1)
System.out.println("No repeating found!");
else
System.out.println("First repeating is " + arr[index]);
}
}
# Python program to find first repeating element
# using Hash Set in O(n) Time and O(n) Space
import sys
def firstRepeatingElement(arr):
s = set()
# If an element is already present, return it
# else insert it
minEle = sys.maxsize
for i in range(len(arr) - 1, -1, -1):
if arr[i] in s:
minEle = min(minEle, i)
s.add(arr[i])
# If no element repeats
return -1 if minEle == sys.maxsize else minEle
arr = [10, 5, 3, 4, 3, 5, 6]
index = firstRepeatingElement(arr)
if index == -1:
print("No repeating found!")
else:
print("First repeating is", arr[index])
// C# program to find first repeating element
// using Hash Set in O(n) Time and O(n) Space
using System;
using System.Collections.Generic;
class GfG
{
static int firstRepeatingElement(int[] arr)
{
HashSet<int> s = new HashSet<int>();
// If an element is already present, return it
// else insert it
int minEle = int.MaxValue;
for (int i = arr.Length - 1; i >= 0; i--)
{
if (s.Contains(arr[i]))
{
minEle = Math.Min(minEle, i);
}
s.Add(arr[i]);
}
// If no element repeats
return minEle == int.MaxValue ? -1 : minEle;
}
static void Main()
{
int[] arr = {10, 5, 3, 4, 3, 5, 6};
int index = firstRepeatingElement(arr);
if (index == -1)
Console.WriteLine("No repeating found!");
else
Console.WriteLine("First repeating is " + arr[index]);
}
}
// JavaScript program to find first repeating element
// using Hash Set in O(n) Time and O(n) Space
function firstRepeatingElement(arr) {
let s = new Set();
// If an element is already present, return it
// else insert it
let minEle = Number.MAX_SAFE_INTEGER;
for (let i = arr.length - 1; i >= 0; i--) {
if (s.has(arr[i])) {
minEle = Math.min(minEle, i);
}
s.add(arr[i]);
}
// If no element repeats
return minEle === Number.MAX_SAFE_INTEGER ? -1 : minEle;
}
let arr = [10, 5, 3, 4, 3, 5, 6];
let index = firstRepeatingElement(arr);
if (index === -1)
console.log("No repeating found!");
else
console.log("First repeating is " + arr[index]);
Output
First repeating is 3
