Count subarrays with similarity number more than K

Question

Similarity number for two arrays X and Y, each with size N, is defined as the number of pairs of indices (i,j) such that X[i]=Y[j] , for 1<=i,j

Now we are given two arrays, of size N and M. We need to find the number of sub arrays of equal sizes from these two arrays such that the similairty number of each subarray pair is greater or equal to given number K.

Example, say we have N=3, M=3, K=1 and arrays be [1,3,4] and [1,5,3] then here answer is 6

Explanation :

({1},{1})
({3},{3})
({1,3},{1,5})
({1,3},{5,3})
({3,4},{5,3})
({1,3,4},{1,5,3})

so ans = 6

How to solve it for given arrays of size N,M and given integer K. Number of elements can't be more than 2000. K is also less than N*M

Approach : Form all subarrays from array 1 of size N, those will be N*(N+1)/2 And same for array 2 of size M. Then try to find similarity number between each subarray pair. But this is very unoptimised way of doing it. What can be better way to solve this problem ? I think Dynamic programming can be used to solve this. Any suggestions ?

For {1,1,2} and {1,1,3} and K=1

{[1(1)],[1(1)]}
{[1(1)],[1(2)]}
{[1(2)],[1(1)]}
{[1(2)],[1(2)]}
{[1(1),1(2)],[1(1)]}
{[1(1),1(2)],[1(2)]}
{[1(1)],[1(1),1(2)]}
{[1(2)],[1(1),1(2)]}
{[1(1),1(2)],[1(1),1(2)]}
{[1(2),2],[1(2),3]}
{[1(1),1(2),2],[1(1),1(2),3]}

Answer 1

Since the contest is now over, just for the sake of completeness, here's my understanding of the editorial answer there (from which I learned a lot). Let's say we had an O(1) time method to calculate the similarity of two contiguous subarrays, one from each array, of length l . Then, for each pair of indexes, (i, j) , we could binary search the smallest l (extending, say to their left) that satisfies similarity k . (Once we have the smallest l , we know that any greater such l also has enough similarity and we can add those counts in O(1) time.) The total time in this case would be O(M * N * log(max (M,N)) .

Well, it turns out there is a way to calculate the similarity of two contiguous subarrays in O(1) : matrix prefix-sums. In a matrix, A ; where each entry, A(i,j) , is 1 if the first array's i th element equals the second array's j th element and 0 otherwise; the sum of the elements in A in the rectangle A(il, jl) , A(i,j) (top-left to bottom-right) is exactly that. And we can calculate that sum in O(1) time with matrix prefix-sums, given O(M*N) preprocessing.