每个元素的频率等于 K 的最长子阵列的长度

给定一个由N 个整数和一个整数K组成的数组arr[] ，任务是找到最长子数组的长度，使得每个元素出现K次。

例子：

Input: arr[] = {3, 5, 2, 2, 4, 6, 4, 6, 5}, K = 2
Output: 8
Explanation: The subarray: {5, 2, 2, 4, 6, 4, 6, 5} of length 8 has frequency of every element as 2.

Input: arr[] = {5, 5, 5, 5}, K = 3
Output: 3
Explanation: The subarray: {5, 5, 5} of length 3 has frequency of every element as 3.

处理方法：按照以下步骤解决问题：

1. 从给定数组生成所有可能的子数组。
2. 对于每个子数组，初始化两个无序映射map1和map2 ，以存储每个元素的频率并存储具有相应频率的元素的计数。
3. 如果对于任何子数组， map2的大小等于1并且当前元素的频率为K ，则意味着每个元素在当前子数组中单独出现K次。
4. 最后，返回所有此类子数组的最大大小。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the length of
// required maximum subarray
int max_subarray_len(int arr[],
                     int N, int K)
{
    // Initialize answer to 0
    int ans = 0;
 
    // Generate all subarrays having i as the
    // starting index and j as the ending index
    for (int i = 0; i < N; i++) {
 
        // Stores frequency of subarray elements
        unordered_map map1;
 
        // Stores subarray elements with
        // respective frequencies
        unordered_map map2;
 
        for (int j = i; j < N; j++) {
 
            // Stores previous
            // frequency of arr[j]
            int prev_freq;
 
            // If arr[j] hasn't
            // occurred previously
            if (map1.find(arr[j])
                == map1.end()) {
 
                // Set frequency as 0
                prev_freq = 0;
            }
 
            else {
 
                // Update previous frequency
                prev_freq = map1[arr[j]];
            }
 
            // Increasing frequency
            // of arr[j] by 1
            map1[arr[j]]++;
 
            // If frequency is stored
            if (map2.find(prev_freq)
                != map2.end()) {
 
                // If previous frequency is 1
                if (map2[prev_freq] == 1) {
 
                    // Rove previous frequency
                    map2.erase(prev_freq);
                }
                else {
 
                    // Decrease previous frequency
                    map2[prev_freq]--;
                }
            }
 
            int new_freq = prev_freq + 1;
 
            // Increment new frequency
            map2[new_freq]++;
 
            // If updated frequency is equal to K
            if (map2.size() == 1
                && (new_freq) == K) {
                ans = max(
                    ans, j - i + 1);
            }
        }
    }
 
    // Return the maximum size
    // of the subarray
    return ans;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 3, 5, 2, 2, 4,
                  6, 4, 6, 5 };
 
    int K = 2;
 
    // Size of Array
    int N = sizeof(arr)
            / sizeof(arr[0]);
 
    // Function Call
    cout << max_subarray_len(
        arr, N, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG
{
 
// Function to find the length of
// required maximum subarray
static int max_subarray_len(int arr[],
                     int N, int K)
{
    // Initialize answer to 0
    int ans = 0;
 
    // Generate all subarrays having i as the
    // starting index and j as the ending index
    for (int i = 0; i < N; i++)
    {
 
        // Stores frequency of subarray elements
        HashMap map1 = new HashMap<>();
 
        // Stores subarray elements with
        // respective frequencies
        HashMap map2 = new HashMap<>();
 
        for (int j = i; j < N; j++)
        {
 
            // Stores previous
            // frequency of arr[j]
            int prev_freq = 0;
 
            // If arr[j] hasn't
            // occurred previously
            if (!map1.containsKey(arr[j]))
            {
 
                // Set frequency as 0
                prev_freq = 0;
            }
 
            else
            {
 
                // Update previous frequency
                prev_freq = map1.get(arr[j]);
            }
 
            // Increasing frequency
            // of arr[j] by 1
            if(map1.containsKey(arr[j]))
            {
                map1.put(arr[j], map1.get(arr[j]) + 1);
            }
            else
            {
                map1.put(arr[j], 1);
            }
 
            // If frequency is stored
            if (map2.containsKey(prev_freq))
            {
 
                // If previous frequency is 1
                if (map2.get(prev_freq) == 1)
                {
 
                    // Rove previous frequency
                    map2.remove(prev_freq);
                }
                else
                {
 
                    // Decrease previous frequency
                    map2.put(prev_freq, map2.get(prev_freq)-1);
                     
                     
                }
            }
 
            int new_freq = prev_freq + 1;
 
            // Increment new frequency
            if(map2.containsKey(new_freq))
            {
                map2.put(new_freq, map2.get(new_freq) + 1);
            }
            else{
                map2.put(new_freq, 1);
            }
 
            // If updated frequency is equal to K
            if (map2.size() == 1
                && (new_freq) == K) {
                ans = Math.max(
                    ans, j - i + 1);
            }
        }
    }
 
    // Return the maximum size
    // of the subarray
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    // Given array arr[]
    int arr[] = { 3, 5, 2, 2, 4,
                  6, 4, 6, 5 };
 
    int K = 2;
 
    // Size of Array
    int N = arr.length;
 
    // Function Call
    System.out.print(max_subarray_len(
        arr, N, K));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above
# approach
from collections import defaultdict
 
# Function to find the length of
# required maximum subarray
def max_subarray_len(arr, N, K):
 
    # Initialize answer to 0
    ans = 0
 
    # Generate all subarrays having
    # i as the starting index and j
    # as the ending index
    for i in range(N):
 
        # Stores frequency of subarray
        # elements
        map1 = defaultdict(int)
 
        # Stores subarray elements with
        # respective frequencies
        map2 = defaultdict(int)
 
        for j in range(i, N):
 
            # If arr[j] hasn't
            # occurred previously
            if (arr[j] not in map1):
 
                # Set frequency as 0
                prev_freq = 0
 
            else:
 
                # Update previous frequency
                prev_freq = map1[arr[j]]
 
            # Increasing frequency
            # of arr[j] by 1
            map1[arr[j]] += 1
 
            # If frequency is stored
            if prev_freq in map2:
 
                # If previous frequency is 1
                if (map2[prev_freq] == 1):
 
                    # Rove previous frequency
                    del map2[prev_freq]
 
                else:
 
                    # Decrease previous frequency
                    map2[prev_freq] -= 1
 
            new_freq = prev_freq + 1
 
            # Increment new frequency
            map2[new_freq] += 1
 
            # If updated frequency is equal
            # to K
            if (len(map2) == 1 and
               (new_freq) == K):
                ans = max(ans, j - i + 1)
                 
    # Return the maximum size
    # of the subarray
    return ans
 
# Driver Code
if __name__ == "__main__":
 
    # Given array arr[]
    arr = [3, 5, 2, 2, 4,
           6, 4, 6, 5]
 
    K = 2
 
    # Size of Array
    N = len(arr)
 
    # Function Call
    print(max_subarray_len(
          arr, N, K))
 
# This code is contributed by Chitranayal

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Function to find the length of
// required maximum subarray
static int max_subarray_len(int []arr,
                     int N, int K)
{
    // Initialize answer to 0
    int ans = 0;
 
    // Generate all subarrays having i as the
    // starting index and j as the ending index
    for (int i = 0; i < N; i++)
    {
 
        // Stores frequency of subarray elements
        Dictionary map1 = new Dictionary();
 
        // Stores subarray elements with
        // respective frequencies
        Dictionary map2 = new Dictionary();
 
        for (int j = i; j < N; j++)
        {
 
            // Stores previous
            // frequency of arr[j]
            int prev_freq = 0;
 
            // If arr[j] hasn't
            // occurred previously
            if (!map1.ContainsKey(arr[j]))
            {
 
                // Set frequency as 0
                prev_freq = 0;
            }
 
            else
            {
 
                // Update previous frequency
                prev_freq = map1[arr[j]];
            }
 
            // Increasing frequency
            // of arr[j] by 1
            if(map1.ContainsKey(arr[j]))
            {
                map1[arr[j]] = map1[arr[j]] + 1;
            }
            else
            {
                map1.Add(arr[j], 1);
            }
 
            // If frequency is stored
            if (map2.ContainsKey(prev_freq))
            {
 
                // If previous frequency is 1
                if (map2[prev_freq] == 1)
                {
 
                    // Rove previous frequency
                    map2.Remove(prev_freq);
                }
                else
                {
 
                    // Decrease previous frequency
                    map2[prev_freq]= map2[prev_freq]-1;
                     
                     
                }
            }
 
            int new_freq = prev_freq + 1;
 
            // Increment new frequency
            if(map2.ContainsKey(new_freq))
            {
                map2[new_freq] = map2[new_freq] + 1;
            }
            else{
                map2.Add(new_freq, 1);
            }
 
            // If updated frequency is equal to K
            if (map2.Count == 1
                && (new_freq) == K) {
                ans = Math.Max(
                    ans, j - i + 1);
            }
        }
    }
 
    // Return the maximum size
    // of the subarray
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    // Given array []arr
    int []arr = { 3, 5, 2, 2, 4,
                  6, 4, 6, 5 };
 
    int K = 2;
 
    // Size of Array
    int N = arr.Length;
 
    // Function Call
    Console.Write(max_subarray_len(
        arr, N, K));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

// Javascript program for the above approach
 
 
// Function to find the length of
// required maximum subarray
function max_subarray_len(arr,N,K)
{
    // Initialize answer to 0
    let ans = 0;
  
    // Generate all subarrays having i as the
    // starting index and j as the ending index
    for (let i = 0; i < N; i++)
    {
  
        // Stores frequency of subarray elements
        let map1 = new Map();
  
        // Stores subarray elements with
        // respective frequencies
        let map2 = new Map();
  
        for (let j = i; j < N; j++)
        {
  
            // Stores previous
            // frequency of arr[j]
            let prev_freq = 0;
  
            // If arr[j] hasn't
            // occurred previously
            if (!map1.has(arr[j]))
            {
  
                // Set frequency as 0
                prev_freq = 0;
            }
  
            else
            {
  
                // Update previous frequency
                prev_freq = map1.get(arr[j]);
            }
  
            // Increasing frequency
            // of arr[j] by 1
            if(map1.has(arr[j]))
            {
                map1.set(arr[j], map1.get(arr[j]) + 1);
            }
            else
            {
                map1.set(arr[j], 1);
            }
  
            // If frequency is stored
            if (map2.has(prev_freq))
            {
  
                // If previous frequency is 1
                if (map2.get(prev_freq) == 1)
                {
  
                    // Rove previous frequency
                    map2.delete(prev_freq);
                }
                else
                {
  
                    // Decrease previous frequency
                    map2.set(prev_freq, map2.get(prev_freq)-1);
                      
                      
                }
            }
  
            let new_freq = prev_freq + 1;
  
            // Increment new frequency
            if(map2.has(new_freq))
            {
                map2.set(new_freq, map2.get(new_freq) + 1);
            }
            else
            {
                map2.set(new_freq, 1);
            }
  
            // If updated frequency is equal to K
            if (map2.size == 1
                && (new_freq) == K)
            {
                ans = Math.max(
                    ans, j - i + 1);
            }
        }
    }
  
    // Return the maximum size
    // of the subarray
    return ans;
}
 
// Driver Code
let arr=[ 3, 5, 2, 2, 4,
                  6, 4, 6, 5 ];
                   
let K = 2;
// Size of Array
let N = arr.length;
 
// Function Call
document.write(max_subarray_len(
arr, N, K));
 
     
 
// This code is contributed by rag2127

输出：

8

时间复杂度： O(N ² )
辅助空间： O(N)

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