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📜  给定数组中最长的置换子序列

📅  最后修改于: 2021-04-21 21:49:42             🧑  作者: Mango

给定包含N个元素的数组arr ,请找到最长子序列的长度,以使其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:

方法:上述问题是在置换子序列上的,因此数组元素的顺序无关紧要,只有重要的是每个元素频率。如果array的长度为N,则排列序列的最大可能长度为N ,最小可能长度为0 。如果长度L的子序列是有效置换,则应存在从1到L的所有元素

  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,并对迭代进行计数,直到观察到0频率为止。如果元素的频率为“ 0”,则返回当前的迭代次数作为所需的长度。

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


输出:
3

时间复杂度: O(N)
辅助空间复杂度: O(N)