给定数组中最长的置换子序列

给定一个包含N 个元素的数组arr ，找到最长子序列的长度，使其成为特定长度的有效排列。如果不存在这样的排列序列，则打印 0。
例子：

Input: arr[] = {3, 2, 1, 6, 5} 
Output: 3 
Explanation: 
Longest permutation subsequence will be [3, 2, 1].
Input: arr[]= {2, 3, 4, 5} 
Output: 0 
Explanation: 
No valid permutation subsequence present as element 1 is missing. 
 

处理方法：上述问题是在置换子序列上，所以数组元素的顺序无关紧要，重要的是每个元素的频率。如果数组的长度为 N ，则置换序列的最大可能长度为N ，最小可能长度为0 。如果长度 L 的子序列是一个有效的排列，那么从 1 到 L 的所有元素都应该存在。

1. 计算数组中 [1, N] 范围内元素的频率
2. 遍历数组中从 1 到 N 的所有元素并计算迭代次数，直到观察到 0 频率。如果元素的频率为“0”，则返回当前迭代次数作为所需长度。

下面是上述方法的实现：

C++

// C++ Program to find length of
// Longest Permutaion Subsequence
// in a given array
 
#include 
using namespace std;
 
// Function to find the
// longest permutation subsequence
int longestPermutation(int a[], int n)
{
 
    // Map data structure to
    // count the frequency of each element
    map freq;
 
    for (int i = 0; i < n; i++) {
 
        freq[a[i]]++;
    }
 
    int len = 0;
 
    for (int i = 1; i <= n; i++) {
 
        // If frequency of element is 0,
        // then we can not move forward
        // as every element should be present
        if (freq[i] == 0) {
            break;
        }
 
        // Increasing the length by one
        len++;
    }
 
    return len;
}
 
// Driver Code
int main()
{
 
    int arr[] = { 3, 2, 1, 6, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << longestPermutation(arr, n)
         << "\n";
 
    return 0;
}

Java

// Java Program to find length of
// Longest Permutaion Subsequence
// in a given array
import java.util.*;
 
class GFG{
  
// Function to find the
// longest permutation subsequence
static int longestPermutation(int arr[], int n)
{
  
    // Map data structure to
    // count the frequency of each element
    HashMap freq = new HashMap();
  
    for (int i = 0; i < n; i++) {
  
        if(freq.containsKey(arr[i])){
            freq.put(arr[i], freq.get(arr[i])+1);
        }else{
            freq.put(arr[i], 1);
        }
    }
  
    int len = 0;
  
    for (int i = 1; i <= n; i++) {
  
        // If frequency of element is 0,
        // then we can not move forward
        // as every element should be present
        if (!freq.containsKey(i)) {
            break;
        }
  
        // Increasing the length by one
        len++;
    }
  
    return len;
}
  
// Driver Code
public static void main(String[] args)
{
  
    int arr[] = { 3, 2, 1, 6, 5 };
    int n = arr.length;
  
    System.out.print(longestPermutation(arr, n));
  
}
}
 
// This code is contributed by Rajput-Ji

C#

// C# Program to find length of
// longest Permutaion Subsequence
// in a given array
 
using System;
using System.Collections.Generic;
 
public class GFG{
 
// Function to find the
// longest permutation subsequence
static int longestPermutation(int []arr, int n)
{
 
    // Map data structure to
    // count the frequency of each element
    Dictionary freq = new Dictionary();
 
    for (int i = 0; i < n; i++) {
 
        if(freq.ContainsKey(arr[i])){
            freq[arr[i]] = freq[arr[i]] + 1;
        }else{
            freq.Add(arr[i], 1);
        }
    }
 
    int len = 0;
 
    for (int i = 1; i <= n; i++) {
 
        // If frequency of element is 0,
        // then we can not move forward
        // as every element should be present
        if (!freq.ContainsKey(i)) {
            break;
        }
 
        // Increasing the length by one
        len++;
    }
 
    return len;
}
 
// Driver Code
public static void Main(String[] args)
{
 
    int []arr = { 3, 2, 1, 6, 5 };
    int n = arr.Length;
 
    Console.Write(longestPermutation(arr, n));
 
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Program to find length of
# Longest Permutaion Subsequence
# in a given array
from collections import defaultdict
 
# Function to find the
# longest permutation subsequence
def longestPermutation(a, n):
  
    # Map data structure to
    # count the frequency of each element
    freq = defaultdict(int)
  
    for i in range( n ):
  
        freq[a[i]] += 1
  
    length = 0
  
    for i in range(1 , n + 1):
  
        # If frequency of element is 0,
        # then we can not move forward
        # as every element should be present
        if (freq[i] == 0):
            break
  
        # Increasing the length by one
        length += 1
         
    return length
  
# Driver Code
if __name__ == "__main__":
  
    arr = [ 3, 2, 1, 6, 5 ]
    n = len(arr)
  
    print(longestPermutation(arr, n))
 
# This code is contributed by chitranayal

Javascript

 
// Javascript Program to find length of
// Longest Permutaion Subsequence
// in a given array
 
// Function to find the
// longest permutation subsequence
function longestPermutation(arr, n)
{
    
    // Map data structure to
    // count the frequency of each element
    let freq = new Map();
    
    for (let i = 0; i < n; i++) {
    
        if(freq.has(arr[i])){
            freq.set(arr[i], freq.get(arr[i])+1);
        }else{
            freq.set(arr[i], 1);
        }
    }
    
    let len = 0;
    
    for (let i = 1; i <= n; i++) {
    
        // If frequency of element is 0,
        // then we can not move forward
        // as every element should be present
        if (!freq.has(i)) {
            break;
        }
    
        // Increasing the length by one
        len++;
    }
    
    return len;
}
  
// Driver code
     
      let arr = [ 3, 2, 1, 6, 5 ];
    let n = arr.length;
    
    document.write(longestPermutation(arr, n));
                                                                                       

输出：

3

时间复杂度： O(N)
辅助空间复杂度： O(N)

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