检查序列是否是两个排列的串联

给定一个包含正整数的数组arr ，任务是检查给定的数组arr是否是两个排列的串联。如果 M 个整数的序列包含从 1 到 M 的所有整数恰好一次，则称为置换。
例子：

Input: arr[] = {1, 2, 5, 3, 4, 1, 1} 
Output: No 
Explanation: 
Given array contains 1 thrice. The first 5 elements form a permutation of length 5, but the remaining 2 elements do not form a permutation.
Input: arr[] = {1, 2, 5, 3, 4, 1, 2} 
Output: Yes 
Explanation: 
Given array arr[] = {1, 2, 5, 3, 4} + {1, 2} 
The first 5 elements form a permutation of length 5 and the remaining 2 elements form a permutation of length 2. 
 

方法：

1. 遍历给定的数组并计算所有元素的总和。
2. 形成一个包含前缀和的前缀数组。
3. 现在，对于 [1, N) 范围内的每个索引
   • 使用以下条件检查从开始到当前索引的元素是否形成排列：

     Sum of K elements = Sum of K natural numbers
     
     where K is the current index
     

• 然后检查剩余元素形成排列。
• 如果是，那么我们打印/返回 Yes。

下面是上述方法的实现：

C++

// C++ program to check if a given sequence
// is a concatenation of two permutations or not
  
#include 
using namespace std;
  
// Function to Check if a given sequence
// is a concatenation of two permutations or not
bool checkPermutation(int arr[], int n)
{
    // Computing the sum of all the
    // elements in the array
    long long sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
  
    // Computing the prefix sum
    // for all the elements in the array
    long long prefix[n + 1] = { 0 };
    prefix[0] = arr[0];
    for (int i = 1; i < n; i++)
        prefix[i] = prefix[i - 1] + arr[i];
  
    // Iterating through the i
    // from lengths 1 to n-1
    for (int i = 0; i < n - 1; i++) {
  
        // Sum of first i+1 elements
        long long lsum = prefix[i];
  
        // Sum of remaining n-i-1 elements
        long long rsum = sum - prefix[i];
  
        // Lengths of the 2 permutations
        long long l_len = i + 1,
                  r_len = n - i - 1;
  
        // Checking if the sums
        // satisfy the formula or not
        if (((2 * lsum)
             == (l_len * (l_len + 1)))
            && ((2 * rsum)
                == (r_len * (r_len + 1))))
            return true;
    }
  
    return false;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 5, 3, 4, 1, 2 };
    int n = sizeof(arr) / sizeof(int);
  
    if (checkPermutation(arr, n))
        cout << "Yes\n";
    else
        cout << "No\n";
  
    return 0;
}

Java

// Java program to check if a given sequence
// is a concatenation of two permutations or not
import java.util.*;
  
class GFG{
  
// Function to Check if a given sequence
// is a concatenation of two permutations or not
static boolean checkPermutation(int []arr, int n)
{
    // Computing the sum of all the
    // elements in the array
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
  
    // Computing the prefix sum
    // for all the elements in the array
    int []prefix = new int[n + 1];
    Arrays.fill(prefix,0);
    prefix[0] = arr[0];
    for (int i = 1; i < n; i++)
        prefix[i] = prefix[i - 1] + arr[i];
  
    // Iterating through the i
    // from lengths 1 to n-1
    for (int i = 0; i < n - 1; i++) {
  
        // Sum of first i+1 elements
        int lsum = prefix[i];
  
        // Sum of remaining n-i-1 elements
        int rsum = sum - prefix[i];
  
        // Lengths of the 2 permutations
        int l_len = i + 1,
                r_len = n - i - 1;
  
        // Checking if the sums
        // satisfy the formula or not
        if (((2 * lsum)
            == (l_len * (l_len + 1)))
            && ((2 * rsum)
                == (r_len * (r_len + 1))))
            return true;
    }
  
    return false;
}
  
// Driver code
public static void main(String args[])
{
    int []arr = { 1, 2, 5, 3, 4, 1, 2 };
    int n = arr.length;
  
    if (checkPermutation(arr, n))
        System.out.println("Yes");
    else
        System.out.println("No");
  
}
}
  
// This code is contributed by Surendra_Gangwar

Python3

# Python program to check if a given sequence
# is a concatenation of two permutations or not
  
# Function to Check if a given sequence
# is a concatenation of two permutations or not
def checkPermutation(arr, n):
      
    # Computing the sum of all the
    # elements in the array
    sum = 0;
    for i in range(n):
        sum += arr[i];
  
    # Computing the prefix sum
    # for all the elements in the array
    prefix = [0]*(n + 1);
    prefix[0] = arr[0];
    for i in range(n):
        prefix[i] = prefix[i - 1] + arr[i];
  
    # Iterating through the i
    # from lengths 1 to n-1
    for i in range(n - 1):
  
        # Sum of first i+1 elements
        lsum = prefix[i];
  
        # Sum of remaining n-i-1 elements
        rsum = sum - prefix[i];
  
        # Lengths of the 2 permutations
        l_len = i + 1
        r_len = n - i - 1;
  
        # Checking if the sums
        # satisfy the formula or not
        if (((2 * lsum)== (l_len * (l_len + 1))) and 
            ((2 * rsum)== (r_len * (r_len + 1)))):
            return True;
      
    return False;
  
# Driver code
if __name__=='__main__': 
  
    arr = [ 1, 2, 5, 3, 4, 1, 2 ]
    n = len(arr)
  
    if (checkPermutation(arr, n)):
        print("Yes");
    else:
        print("No");
  
# This code is contributed by Princi Singh

C#

// C# program to check if a given sequence
// is a concatenation of two permutations or not
using System;
  
class GFG{
   
// Function to Check if a given sequence
// is a concatenation of two permutations or not
static bool checkPermutation(int []arr, int n)
{
    // Computing the sum of all the
    // elements in the array
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
   
    // Computing the prefix sum
    // for all the elements in the array
    int []prefix = new int[n + 1];
    prefix[0] = arr[0];
    for (int i = 1; i < n; i++)
        prefix[i] = prefix[i - 1] + arr[i];
   
    // Iterating through the i
    // from lengths 1 to n-1
    for (int i = 0; i < n - 1; i++) {
   
        // Sum of first i+1 elements
        int lsum = prefix[i];
   
        // Sum of remaining n-i-1 elements
        int rsum = sum - prefix[i];
   
        // Lengths of the 2 permutations
        int l_len = i + 1,
                r_len = n - i - 1;
   
        // Checking if the sums
        // satisfy the formula or not
        if (((2 * lsum)
            == (l_len * (l_len + 1)))
            && ((2 * rsum)
                == (r_len * (r_len + 1))))
            return true;
    }
   
    return false;
}
   
// Driver code
public static void Main(String []args)
{
    int []arr = { 1, 2, 5, 3, 4, 1, 2 };
    int n = arr.Length;
   
    if (checkPermutation(arr, n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
   
}
}
  
// This code is contributed by Rajput-Ji

Javascript

  
// JavaScript program to check if a given sequence
// is a concatenation of two permutations or not
  
// Function to Check if a given sequence
// is a concatenation of two permutations or not
function checkPermutation(arr, n)
{
    // Computing the sum of all the
    // elements in the array
    let sum = 0;
    for (let i = 0; i < n; i++)
        sum += arr[i];
    
    // Computing the prefix sum
    // for all the elements in the array
    let prefix = Array.from({length: n+1}, (_, i) => 0);
    prefix[0] = arr[0];
    for (let i = 1; i < n; i++)
        prefix[i] = prefix[i - 1] + arr[i];
    
    // Iterating through the i
    // from lengths 1 to n-1
    for (let i = 0; i < n - 1; i++) {
    
        // Sum of first i+1 elements
        let lsum = prefix[i];
    
        // Sum of remaining n-i-1 elements
        let rsum = sum - prefix[i];
    
        // Lengths of the 2 permutations
        let l_len = i + 1,
                r_len = n - i - 1;
    
        // Checking if the sums
        // satisfy the formula or not
        if (((2 * lsum)
            == (l_len * (l_len + 1)))
            && ((2 * rsum)
                == (r_len * (r_len + 1))))
            return true;
    }
    
    return false;
}
   
  
// Driver Code
      
        let arr = [ 1, 2, 5, 3, 4, 1, 2 ];
    let n = arr.length;
    
    if (checkPermutation(arr, n))
        document.write("Yes");
    else
        document.write("No");
                    

输出：

Yes