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 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:
- Initialize a variable j, to store the maximum value of the index such that there is no repeated elements between index i and j
- Traverse the array and keep updating j based on previous occurrence of a[i[ stored in the HashMap.
- After updating j, update ans accordingly to store maximum length of desired subarray.
- 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 |
5
Time Complexity: O(NlogN)
Auxiliary Space: O(N)
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