LeetCode 1940: Longest Common Subsequence Between Sorted Arrays Solution

Master LeetCode problem 1940 (Longest Common Subsequence Between Sorted Arrays), 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.

1940. Longest Common Subsequence Between Sorted Arrays

Medium

Array Hash Table Counting

Problem Explanation

Explanation:

The problem asks us to find the longest common subsequence between two sorted arrays. A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.To solve this problem efficiently, we can use a dynamic programming approach. We will create a 2D array to store the length of the longest common subsequence up to each pair of elements from both arrays.#### Algorithmic Steps:

Initialize a 2D array dp of size n x m, where n is the length of the first array and m is the length of the second array.
Iterate over each pair of elements from both arrays.
If the elements are equal, update dp[i][j] to dp[i-1][j-1] + 1 (length of common subsequence up to the previous pair plus 1).
If the elements are not equal, update dp[i][j] to the maximum of dp[i-1][j] and dp[i][j-1].
The final answer will be dp[n-1][m-1], which represents the length of the longest common subsequence.

Solution Code

class Solution {
    public int longestCommonSubsequence(int[] nums1, int[] nums2) {
        int n = nums1.length;
        int m = nums2.length;
        int[][] dp = new int[n][m];
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (nums1[i] == nums2[j]) {
                    dp[i][j] = (i > 0 && j > 0) ? dp[i-1][j-1] + 1 : 1;
                } else {
                    int prev1 = (i > 0) ? dp[i-1][j] : 0;
                    int prev2 = (j > 0) ? dp[i][j-1] : 0;
                    dp[i][j] = Math.max(prev1, prev2);
                }
            }
        }
        
        return dp[n-1][m-1];
    }
}

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