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.
Input: A = [1, 2, 3], B = [2, 3, 1]
The common subarrays are , , , [2, 3]
Hence, total count = 4
Input: A = [1, 2, 3, 4, 5], B = [2, 3, 1, 4, 5]
The common subarrays are , , , , , [2, 3], [4, 5]
Hence, total count = 7
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).
The idea is to use Hashing to solve this problem efficiently.
- Create a Hash array H of size N+1.
- Represent all elements of A by their respective indices:
Element Representaion A 0 A 1 A 2 . . and so on.
- Use array H to store this representation, H[ A[ i ] ] = i
- Update the elements of B according to this new representation using H, B[ i ] = H[ B[ i ] ]
- 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:
A = [4, 3, 1, 2, 5]
B = [3, 1, 2, 4, 5]
Common subarrays are , , , , , [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  of length K = 1
Element at index 4 forms a subarray  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:
Time complexity: O(N)
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