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Longest Subarray consisiting of unique elements from an Array
  • Difficulty Level : Hard
  • Last Updated : 21 Aug, 2020

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:
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:
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 using HashSet. Find the longest subarray satisfying the condition. 



Time Complexity: O(N3logN) 
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimised using HashMap. Follow the steps below to solve the problem:

  1. Initialize a variable j, to store the maximum value of the index such that there is no repeated elements between index i and j
  2. Traverse the array and keep updating j based on previous occurrence of a[i[ stored in the HashMap.
  3. After updating j, update ans accordingly to store maximum length of desired subarray.
  4. Print ans, after traversal is completed.

Below is the implementation of above approach:

C++




// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find largest
// subarray with no dublicates
int 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 Code
int32_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




// Java program to implement
// the above approach
import java.util.*;
class GFG{
  
// Function to find largest
// subarray with no dublicates
static 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 Code
public 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




# Python3 program to implement
# the above approach
from collections import defaultdict
  
# Function to find largest
# subarray with no dublicates
def 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 Code
arr = [ 1, 2, 3, 4, 5, 1, 2, 3 ]
n = len(arr)
  
# Function call
print(largest_subarray(arr, n))
  
# This code is contributed by Shivam Singh

C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
  
class GFG{
  
// Function to find largest
// subarray with no dublicates
static 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 Code
public 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
Output: 
5

Time Complexity: O(NlogN) 
Auxiliary Space: O(N)

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