Longest Subarray consisting of unique elements from an Array

09 Jun, 2022
Given an array arr[] consisting of N integers, the task is to find the largest subarray consisting of unique elements only.
Examples:
Input: arr[] = {1, 2, 3, 4, 5, 1, 2, 3} Output: 5 Explanation: One possible subarray is {1, 2, 3, 4, 5}.
Input: arr[]={1, 2, 4, 4, 5, 6, 7, 8, 3, 4, 5, 3, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4} Output: 8 Explanation: Only possible subarray is {3, 4, 5, 6, 7, 8, 1, 2}.
Naive Approach: The simplest approach to solve the problem is to generate all subarrays from the given array and check if it contains any duplicates or not to use HashSet. Find the longest subarray satisfying the condition. Time Complexity: O(N3logN) Auxiliary Space: O(N)
Efficient Approach: The above approach can be optimized by HashMap. Follow the steps below to solve the problem:
Initialize a variable j, to store the maximum value of the index such that there are no repeated elements between index i and jTraverse the array and keep updating j based on the previous occurrence of a[i] stored in the HashMap.After updating j, update ans accordingly to store the maximum length of the desired subarray.Print ans, after traversal, is completed.
Initialize a variable j, to store the maximum value of the index such that there are no repeated elements between index i and j
Traverse the array and keep updating j based on the previous occurrence of a[i] stored in the HashMap.
After updating j, update ans accordingly to store the maximum length of the desired subarray.
Print ans, after traversal, is completed.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std; // Function to find largest// subarray with no duplicatesint largest_subarray(int a[], int n){    // Stores index of array elements    unordered_map<int, int> index;    int ans = 0;    for (int i = 0, j = 0; i < n; i++) {         // Update j based on previous        // occurrence of a[i]        j = max(index[a[i]], j);         // Update ans to store maximum        // length of subarray        ans = max(ans, i - j + 1);         // Store the index of current        // occurrence of a[i]        index[a[i]] = i + 1;    }     // Return final ans    return ans;} // Driver Codeint32_t main(){    int arr[] = { 1, 2, 3, 4, 5, 1, 2, 3 };    int n = sizeof(arr) / sizeof(arr[0]);    cout << largest_subarray(arr, n);}
// Java program to implement// the above approachimport java.util.*;class GFG{ // Function to find largest// subarray with no duplicatesstatic int largest_subarray(int a[], int n){    // Stores index of array elements    HashMap<Integer,            Integer> index = new HashMap<Integer,                                         Integer>();    int ans = 0;    for(int i = 0, j = 0; i < n; i++)    {         // Update j based on previous        // occurrence of a[i]        j = Math.max(index.containsKey(a[i]) ?                             index.get(a[i]) : 0, j);         // Update ans to store maximum        // length of subarray        ans = Math.max(ans, i - j + 1);         // Store the index of current        // occurrence of a[i]        index.put(a[i], i + 1);    }     // Return final ans    return ans;} // Driver Codepublic static void main(String[] args){    int arr[] = { 1, 2, 3, 4, 5, 1, 2, 3 };    int n = arr.length;    System.out.print(largest_subarray(arr, n));}} // This code is contributed by Rajput-Ji
# Python3 program to implement# the above approachfrom collections import defaultdict # Function to find largest# subarray with no duplicatesdef largest_subarray(a, n):     # Stores index of array elements    index = defaultdict(lambda : 0)         ans = 0    j = 0     for i in range(n):         # Update j based on previous        # occurrence of a[i]        j = max(index[a[i]], j)         # Update ans to store maximum        # length of subarray        ans = max(ans, i - j + 1)         # Store the index of current        # occurrence of a[i]        index[a[i]] = i + 1         i += 1     # Return final ans    return ans # Driver Codearr = [ 1, 2, 3, 4, 5, 1, 2, 3 ]n = len(arr) # Function callprint(largest_subarray(arr, n)) # This code is contributed by Shivam Singh
// C# program to implement// the above approachusing System;using System.Collections.Generic; class GFG{ // Function to find largest// subarray with no duplicatesstatic int largest_subarray(int []a, int n){         // Stores index of array elements    Dictionary<int,               int> index = new Dictionary<int,                                           int>();    int ans = 0;    for(int i = 0, j = 0; i < n; i++)    {         // Update j based on previous        // occurrence of a[i]        j = Math.Max(index.ContainsKey(a[i]) ?                                 index[a[i]] : 0, j);         // Update ans to store maximum        // length of subarray        ans = Math.Max(ans, i - j + 1);         // Store the index of current        // occurrence of a[i]        if(index.ContainsKey(a[i]))            index[a[i]] = i + 1;        else            index.Add(a[i], i + 1);    }     // Return readonly ans    return ans;} // Driver Codepublic static void Main(String[] args){    int []arr = { 1, 2, 3, 4, 5, 1, 2, 3 };    int n = arr.Length;         Console.Write(largest_subarray(arr, n));}} // This code is contributed by Amit Katiyar
<script> // Javascript program to implement// the above approach // Function to find largest// subarray with no duplicatesfunction largest_subarray(a, n){    // Stores index of array elements    let index = new Map();    let ans = 0;    for(let i = 0, j = 0; i < n; i++)    {          // Update j based on previous        // occurrence of a[i]        j = Math.max(index.has(a[i]) ?                             index.get(a[i]) : 0, j);          // Update ans to store maximum        // length of subarray        ans = Math.max(ans, i - j + 1);          // Store the index of current        // occurrence of a[i]        index.set(a[i], i + 1);    }      // Return final ans    return ans;} // Driver code     let arr = [ 1, 2, 3, 4, 5, 1, 2, 3 ];    let n = arr.length;    document.write(largest_subarray(arr, n)); </script>
5
 
Time Complexity: O(N) in best case and O(n^2) in worst case.
NOTE: We can make Time complexity equal to O(n * logn) by using balanced binary tree structures (`std::map` in c++ and `TreeMap` in Java.) instead of Hash structures.Auxiliary Space: O(N)
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surindertarika1234
anikaseth98
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frequency-counting
Java-HashMap
subarray
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