1到N的两个不同排列中的公共子数组的计数

给定两个具有相同长度N的数组A和B ，并填充了从1到N的自然数的排列，任务是计算A和B中公共子数组的数量。
例子：

Input: A = [1, 2, 3], B = [2, 3, 1] 
Output: 4 
Explanation: 
The common subarrays are [1], [2], [3], [2, 3] 
Hence, total count = 4
Input: A = [1, 2, 3, 4, 5], B = [2, 3, 1, 4, 5] 
Output: 7 
Explanation: 
The common subarrays are [1], [2], [3], [4], [5], [2, 3], [4, 5] 
Hence, total count = 7 
 

天真的方法：
想法是分别生成A和B的所有子阵列，每个子阵列将采用O(N ² ) 。现在，将A的所有子阵列与B的所有子阵列进行比较，并计算公共子阵列。这需要O(N ⁴ ) 。
高效方法：
这个想法是使用哈希来有效地解决这个问题。

1. 创建一个大小为N + 1的哈希数组H。
2. 用它们各自的索引表示A的所有元素：

Element      Representaion
A[0]      0
A[1]      1
A[2]      2
.
.
and so on.

2. 使用数组H来存储此表示， H [A [i]] = i
3. 使用H根据此新表示更新B的元素， B [i] = H [B [i]]
4. 现在，数组A可以表示为[0，1，2，..N]，因此只需对B中具有连续元素的子数组的数量进行计数即可。一旦我们获得了连续元素的子数组的长度K ，就可以使用以下关系式计算总的可能子数组：

Total number of subarrays = (K * (K + 1)) / 2 
 

查看此示例以详细了解此方法：

Example: 
A = [4, 3, 1, 2, 5] 
B = [3, 1, 2, 4, 5] 
Common subarrays are [1], [2], [3], [4], [5], [3, 1], [1, 2], [3, 1, 2] = 8
1. Represent A[i] as i, and store in H as H[A[i]] = i, Now array H from index 1 to N is, 
H = [2, 3, 1, 0, 4]
2. Update B according to H, B[i] = H[B[i]] 
B = [1, 2, 3, 0, 4]
3. Look for subarray in B with consecutive elements, 
Subarray from index 0 to 2 is [1, 2, 3], consisting of consecutive elements with length K = 3 
Element at index 3 forms a subarray [0] of length K = 1 
Element at index 4 forms a subarray [4] of length K = 1
4. Total number of common subarrays = 
(3*(3+1))/2 + (1*(1+1))/2 + (1*(1+1))/2 = 6 + 1 + 1 = 8 
 

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include
using namespace std;
 
int commonSubarrays(int *A, int *B, int N)
{
    // Initialising Map for
    // Index Mapping
    int Map[N + 1];
 
    // Mapping elements of A
    for(int i = 0 ; i< N; i++)
        Map[*(A + i)] = i;
 
    // Modify elements of B
    // according to Map
    for (int i = 0; i < N; i++)
    {
        // Changing B[i] as
        // the index of B[i] in A
        *(B + i) = Map[*(B + i)];
    }
     
    // Count of common subarrays
    int count = 0;
 
    // Traversing array B
    int i = 0, K;
    while (i < N)
    {
        K = 1;
        i+= 1;
 
        // While consecutive elements
        // are found, we increment K
        while (i < N && B[i] == B[i - 1] + 1)
        {
            i += 1;
            K += 1;
        }
         
        // Add number of subarrays
        //with length K
        // to total count
        count = count + ((K) * (K + 1)) / 2;
    }
    return count;
}
 
// Driver code
int main()
{
    int N = 3;
    int A[] = {1, 2, 3};
    int B[] = {2, 3, 1};
    cout << (commonSubarrays(A, B, N))
         << endl;
 
    N = 5;
    int C[] = {1, 2, 3, 4, 5};
    int D[] = {2, 3, 1, 4, 5};
    cout << (commonSubarrays(C, D, N));
}
 
// This code is contributed by chitranayal

Java

// Java implementation of the above approach
class GFG{
 
static int commonSubarrays(int []A,
                           int []B, int N)
{
     
    // Initialising Map for
    // Index Mapping
    int []Map = new int[N + 1];
 
    // Mapping elements of A
    for(int i = 0; i< N; i++)
       Map[A[i]] = i;
 
    // Modify elements of B
    // according to Map
    for(int i = 0; i < N; i++)
    {
        
       // Changing B[i] as
       // the index of B[i] in A
       B[i] = Map[B[i]];
    }
     
    // Count of common subarrays
    int count = 0;
 
    // Traversing array B
    int i = 0, K;
    while (i < N)
    {
        K = 1;
        i+= 1;
 
        // While consecutive elements
        // are found, we increment K
        while (i < N && B[i] == B[i - 1] + 1)
        {
            i += 1;
            K += 1;
        }
         
        // Add number of subarrays
        //with length K
        // to total count
        count = count + ((K) * (K + 1)) / 2;
    }
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 3;
    int A[] = {1, 2, 3};
    int B[] = {2, 3, 1};
    System.out.print(commonSubarrays(A, B, N));
    System.out.print("\n");
     
    N = 5;
    int C[] = {1, 2, 3, 4, 5};
    int D[] = {2, 3, 1, 4, 5};
    System.out.print(commonSubarrays(C, D, N));
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 implementation of above approach
 
def commonSubarrays(A, B, N):
 
    # Initialising Map for
    # Index Mapping
    Map = [0 for i in range(N + 1)]
 
    # Mapping elements of A
    for i in range(N):
        Map[A[i]]= i
 
    # Modify elements of B
    # according to Map
    for i in range(N)    :
         
        # Changing B[i] as
        # the index of B[i] in A
        B[i]= Map[B[i]]
 
    # Count of common subarrays
    count = 0
 
    # Traversing array B
    i = 0
    while i

---

C#

// C# implementation of the above approach
using System;
class GFG{
 
static int commonSubarrays(int []A,
                           int []B,
                           int N)
{
     
    // Initialising Map for
    // Index Mapping
    int []Map = new int[N + 1];
 
    // Mapping elements of A
    for(int i = 0; i < N; i++)
       Map[A[i]] = i;
 
    // Modify elements of B
    // according to Map
    for(int i = 0; i < N; i++)
    {
        
       // Changing B[i] as
       // the index of B[i] in A
       B[i] = Map[B[i]];
    }
     
    // Count of common subarrays
    int count = 0;
 
    // Traversing array B
    int a = 0, K;
    while (a < N)
    {
        K = 1;
        a += 1;
 
        // While consecutive elements
        // are found, we increment K
        while (a < N && B[a] == B[a - 1] + 1)
        {
            a += 1;
            K += 1;
        }
         
        // Add number of subarrays
        //with length K
        // to total count
        count = count + ((K) * (K + 1)) / 2;
    }
    return count;
}
 
// Driver code
public static void Main()
{
    int N = 3;
    int []A = {1, 2, 3};
    int []B = {2, 3, 1};
    Console.Write(commonSubarrays(A, B, N));
    Console.Write("\n");
     
    N = 5;
    int []C = {1, 2, 3, 4, 5};
    int []D = {2, 3, 1, 4, 5};
    Console.Write(commonSubarrays(C, D, N));
}
}
 
// This code is contributed by Code_Mech

Javascript

// Javascript implementation of the above approach
 
function commonSubarrays(A, B, N)
{
       
    // Initialising Map for
    // Index Mapping
    let Map = Array.from({length: N+1}, (_, i) => 0);
   
    // Mapping elements of A
    for(let i = 0; i< N; i++)
       Map[A[i]] = i;
   
    // Modify elements of B
    // according to Map
    for(let i = 0; i < N; i++)
    {
          
       // Changing B[i] as
       // the index of B[i] in A
       B[i] = Map[B[i]];
    }
       
    // Count of common subarrays
    let count = 0;
   
    // Traversing array B
    let i = 0, K;
    while (i < N)
    {
        K = 1;
        i+= 1;
   
        // While consecutive elements
        // are found, we increment K
        while (i < N && B[i] == B[i - 1] + 1)
        {
            i += 1;
            K += 1;
        }
           
        // Add number of subarrays
        //with length K
        // to total count
        count = count + ((K) * (K + 1)) / 2;
    }
    return count;
}
 
  // Driver Code
    let N = 3;
    let A = [1, 2, 3];
    let B = [2, 3, 1];
    document.write(commonSubarrays(A, B, N));
    document.write("<br/>");
       
    N = 5;
    let C = [1, 2, 3, 4, 5];
    let D = [2, 3, 1, 4, 5];
    document.write(commonSubarrays(C, D, N));  
      
    // This code is contributed by target_2.

输出：

4
7

时间复杂度： O(N)

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