Sponsored
Sponsored
This approach uses a dynamic programming (DP) solution where we maintain a 2D array called dp. The element dp[i][j] represents the maximum number of lines that can be drawn without crossing up to the i-th element of nums1 and the j-th element of nums2.
The DP formulation is based on: if nums1[i] == nums2[j], then dp[i][j] = dp[i-1][j-1] + 1. Otherwise, dp[i][j] = max(dp[i-1][j], dp[i][j-1]).
The final answer is found in dp[m][n], where m and n are the lengths of nums1 and nums2 respectively. This approach runs in O(m * n) time complexity and uses O(m * n) space.
Time Complexity: O(m * n), where m and n are the lengths of nums1 and nums2.
Space Complexity: O(m * n), for the dp array.
1using System;
2
3public class Solution {
4 public int MaxUncrossedLines(int[] A, int[] B) {
5 int m = A.Length, n = B.Length;
6 int[,] dp = new int[m + 1, n + 1];
7 for (int i = 1; i <= m; i++) {
8 for (int j = 1; j <= n; j++) {
9 if (A[i - 1] == B[j - 1])
10 dp[i, j] = dp[i - 1, j - 1] + 1;
11 else
12 dp[i, j] = Math.Max(dp[i - 1, j], dp[i, j - 1]);
13 }
14 }
15 return dp[m, n];
16 }
17
18 public static void Main() {
19 Solution solution = new Solution();
20 int[] nums1 = {1, 4, 2};
21 int[] nums2 = {1, 2, 4};
22 Console.WriteLine("Output: " + solution.MaxUncrossedLines(nums1, nums2));
23 }
24}In this C# solution, a 2D array dp is employed. The nested loops iterate over every pair of indices from the sequences, updating the state of each dp[i, j] similarly to the previous implementations. By utilizing this dynamic programming approach, the maximum non-intersecting lines can be determined for the entire range of both sequences.
This approach is an optimization of the basic dynamic programming solution by reducing space complexity. Instead of utilizing a 2D array, we can use a single-dimensional array to store only the current and previous row information, thus reducing the space usage. This takes advantage of the fact that the current DP state depends only on the previous state, not all preceding states.
We maintain two 1D arrays, and they get updated in each iteration over the second array, resulting in reduced space complexity to O(min(m, n)).
Time Complexity: O(m * n), where m and n are the lengths of nums1 and nums2.
Space Complexity: O(min(m, n)), for the dp array.
By leveraging a single array dp of size equal to nums2, the solution benefits from reduced space usage. We keep track of the previous computed value using a temporary variable prev. For each element pair, we determine the longest line by either incrementing based on a match or preserving the maximum from prior positions.