Introduction Link to heading
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
Note:
1 <= A.length <= 500
1 <= B.length <= 500
1 <= A[i], B[i] <= 2000
Solution Link to heading
This problem can be transformed in to longest common sequence.
func maxUncrossedLines(A []int, B []int) int {
n, m := len(A), len(B)
dp := make([][]int, n+1)
dp[0] = make([]int, m+1)
for i := 1; i <= n; i++ {
dp[i] = make([]int, m+1)
for j := 1; j <= m; j++ {
if A[i-1] == B[j-1] {
dp[i][j] = dp[i-1][j-1] + 1
} else {
dp[i][j] = max(dp[i][j-1], dp[i-1][j])
}
}
}
return dp[n][m]
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
Performance Link to heading
Current performance
Runtime: 4 ms
Memory Usage: 6.1 MB
We do not need to keep the full 2D array, just a previous row and current.
func maxUncrossedLines(A []int, B []int) int {
n, m := len(A), len(B)
prev, curr := make([]int, m+1), make([]int, m+1)
for i := 1; i <= n; i++ {
for j := 1; j <= m; j++ {
if A[i-1] == B[j-1] {
curr[j] = prev[j-1] + 1
} else {
curr[j] = max(curr[j-1], prev[j])
}
}
prev, curr = curr, prev
}
return prev[m]
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
This solution is faster:
Runtime: 0 ms
Memory Usage: 2.3 MB