由数组中的唯一元素组成的最长子数组

给定一个由N 个整数组成的数组arr[] ，任务是找到仅由唯一元素组成的最大子数组。

例子：

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}.

朴素方法：解决问题的最简单方法是从给定数组生成所有子数组，并检查它是否包含任何重复项或不使用 HashSet。找到满足条件的最长子数组。
时间复杂度： O(N ³ logN)
辅助空间： O(N)

高效方法：上述方法可以通过HashMap进行优化。请按照以下步骤解决问题：

1. 初始化一个变量j ，存储索引的最大值，使得索引i 和 j之间没有重复元素
2. 遍历数组并根据存储在 HashMap 中的先前出现的 a[i] 不断更新j 。
3. 更新j 后，相应地更新ans以存储所需子数组的最大长度。
4. 遍历完成后打印ans 。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
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 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 index = new HashMap();
    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 index = new Dictionary();
    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

Javascript

 
// Javascript program to implement
// the above approach
 
// Function to find largest
// subarray with no dublicates
function 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));
 

输出：

5

时间复杂度： O(NlogN)
辅助空间： O(N)

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