LeetCode 1035: Uncrossed Lines Solution

Master LeetCode problem 1035 (Uncrossed Lines), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1035. Uncrossed Lines

Medium

Array Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We can define a 2D array dp where dp[i][j] represents the maximum number of connecting lines we can draw using the first i elements of nums1 and the first j elements of nums2. We can then iterate through the arrays and update the dp array based on whether the current elements are equal or not.

Here are the steps:

Initialize a 2D array dp with dimensions nums1.length + 1 and nums2.length + 1.
Iterate through nums1 and nums2 using two nested loops.
If nums1[i-1] == nums2[j-1], update dp[i][j] as dp[i-1][j-1] + 1.
Otherwise, update dp[i][j] as the maximum of dp[i-1][j] and dp[i][j-1].
Finally, return dp[nums1.length][nums2.length].

The time complexity of this solution is O(m * n) where m and n are the lengths of nums1 and nums2 respectively. The space complexity is also O(m * n) for the dp array.

Solution Code

class Solution {
    public int maxUncrossedLines(int[] nums1, int[] nums2) {
        int m = nums1.length;
        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++) {
                if (nums1[i - 1] == nums2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        
        return dp[m][n];
    }
}

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