Skip to content

1035. Uncrossed Lines 👍

  • Time: $O(mn)$
  • Space: $O(mn)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
class Solution {
 public:
  int maxUncrossedLines(vector<int>& nums1, vector<int>& nums2) {
    const int m = nums1.size();
    const int n = nums2.size();
    vector<vector<int>> dp(m + 1, vector<int>(n + 1));

    for (int i = 1; i <= m; ++i)
      for (int j = 1; j <= n; ++j)
        dp[i][j] = nums1[i - 1] == nums2[j - 1]
                       ? dp[i - 1][j - 1] + 1
                       : max(dp[i - 1][j], dp[i][j - 1]);

    return dp[m][n];
  }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
class Solution {
  public int maxUncrossedLines(int[] nums1, int[] nums2) {
    final int m = nums1.length;
    final int n = nums2.length;
    int[][] dp = new int[m + 1][n + 1];

    for (int i = 1; i <= m; ++i)
      for (int j = 1; j <= n; ++j)
        dp[i][j] = nums1[i - 1] == nums2[j - 1] ? dp[i - 1][j - 1] + 1
                                                : Math.max(dp[i - 1][j], dp[i][j - 1]);

    return dp[m][n];
  }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
class Solution:
  def maxUncrossedLines(self, nums1: list[int], nums2: list[int]) -> int:
    m = len(nums1)
    n = len(nums2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]

    for i in range(1, m + 1):
      for j in range(1, n + 1):
        dp[i][j] = (dp[i - 1][j - 1] + 1
                    if nums1[i - 1] == nums2[j - 1]
                    else max(dp[i - 1][j], dp[i][j - 1]))

    return dp[m][n]