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Count of common subarrays in two different permutations of 1 to N

  • Last Updated : 17 Aug, 2021

Given two arrays A and B of the same length N, filled with a permutation of natural numbers from 1 to N, the task is to count the number of common subarrays in A and B.
Examples:
 

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

 

Naive Approach: 
The idea is to generate all subarrays of A and B separately, which would take O(N2) for each array. Now, compare all subarrays of A with all subarrays of B and count common subarrays. It would take O(N4).
Efficient Approach: 
The idea is to use Hashing to solve this problem efficiently. 
 

  1. Create a Hash array H of size N+1.
  2. Represent all elements of A by their respective indices:
     
Element      Representation
A[0]      0
A[1]      1
A[2]      2
.
.
and so on.

       3. Use array H to store this representation, H[ A[ i ] ] = i

       4. Update the elements of B according to this new representation using H, B[ i ] = H[ B[ i ] ]

       5. Now, array A can be represented as [0, 1, 2, ..N], so simply count number of subarrays in B which have consecutive elements. Once we get length K of subarray of consecutive elements, count total possible subarray using following relation: 
 

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

Look at this example to understand this approach in detail:
 

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 
 

Below is the implementation of the above approach:
 

C++




// C++ implementation of above approach
#include<bits/stdc++.h>
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<N:
        K = 1
        i+= 1
 
        # While consecutive elements
        # are found, we increment K
        while i<N and 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
N = 3
A =[1, 2, 3]
B =[2, 3, 1]
print(commonSubarrays(A, B, N))
 
N = 5
A =[1, 2, 3, 4, 5]
B =[2, 3, 1, 4, 5]
print(commonSubarrays(A, B, N))

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




<script>
// 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.
</script>
Output: 
4
7

 

Time complexity: O(N)  
Auxiliary Space: O(N)


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