Problem

We write the integers of A and B (in the order they are given) on two separate horizontal lines.

Now, we may draw connecting lines: a straight line connecting two numbers A[i] and B[j] such that:

A[i] == B[j]; The line we draw does not intersect any other connecting (non-horizontal) line. Note that a connecting lines cannot intersect even at the endpoints: each number can only belong to one connecting line.

Return the maximum number of connecting lines we can draw in this way.

example 1

Input: A = [1,4,2], B = [1,2,4]
Output: 2
Explanation: We can draw 2 uncrossed lines as in the diagram.
We cannot draw 3 uncrossed lines, because the line from A[1]=4 to B[2]=4 will intersect the line from A[2]=2 to B[1]=2.

example 2

Input: A = [2,5,1,2,5], B = [10,5,2,1,5,2]
Output: 3

example 3

Input: A = [1,3,7,1,7,5], B = [1,9,2,5,1]
Output: 2

Constraints

• 1 <= len(A) <= 500
• 1 <= len(B) <= 500
• 1 <= A[i], B[i] <= 2000

Solution

Dynamic programming

Lets describe our dp function.

i – iterator over A

j – iterator over B

• If i<0 or j<0 – return 0
• Compare A[i] and B[j] :
  • if equal – run dp with i-1, j-1; return with + 1
  • if not equeal – return max of dp(i-1, j) and dp(i, j-1)

func maxUncrossedLines(A []int, B []int) int {
    m := make(map[int]map[int]int, len(A))
    for idx:=0;idx<len(A);idx++{
        m[idx] = make(map[int]int, len(B))
    }
    return dp(A,B,len(A)-1,len(B)-1,m)
}

func dp(A,B []int, i,j int,m map[int]map[int]int) int {
    if i == -1 || j==-1 {
        return 0
    }
    if _, ok :=m[i][j]; ok {
        return m[i][j]
    }
    if A[i] == B[j] {
        res :=  dp(A,B,i-1,j-1,m)+1
        m[i][j]=res
        return res
    } 
        l := dp(A,B,i,j-1,m)
        r:=dp(A,B,i-1,j,m)
    res:= max(l,r)
    m[i][j]=res
    return res
    
}

func max(a,b int) int {
    if a>b {
        return a   
    }
    return b
}