LeetCode-in-Java

1143. Longest Common Subsequence

Medium

Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0.

A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

• For example, "ace" is a subsequence of "abcde".

A common subsequence of two strings is a subsequence that is common to both strings.

Example 1:

Input: text1 = “abcde”, text2 = “ace”

Output: 3

Explanation: The longest common subsequence is “ace” and its length is 3.

Example 2:

Input: text1 = “abc”, text2 = “abc”

Output: 3

Explanation: The longest common subsequence is “abc” and its length is 3.

Example 3:

Input: text1 = “abc”, text2 = “def”

Output: 0

Explanation: There is no such common subsequence, so the result is 0.

Constraints:

• 1 <= text1.length, text2.length <= 1000
• text1 and text2 consist of only lowercase English characters.

Solution

public class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int n = text1.length();
        int m = text2.length();
        int[][] dp = new int[n + 1][m + 1];
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (text1.charAt(i - 1) == text2.charAt(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[n][m];
    }
}

Time Complexity (Big O Time):

The program uses dynamic programming to fill in a 2D array dp of dimensions (n+1) x (m+1), where n is the length of text1, and m is the length of text2. It iterates through this 2D array using nested loops:

• The outer loop runs n times, where n is the length of text1.
• The inner loop runs m times, where m is the length of text2.

Inside the nested loops, each iteration involves simple constant-time operations (comparisons, assignments, and max calculations). Therefore, the overall time complexity of the program is O(n * m), where ‘n’ and ‘m’ are the lengths of the input strings text1 and text2, respectively.

Space Complexity (Big O Space):

The program uses a 2D array dp of dimensions (n+1) x (m+1) to store intermediate results for dynamic programming. As such, the space complexity is determined by the size of this array:

• The number of rows in dp is (n+1) where ‘n’ is the length of text1.
• The number of columns in dp is (m+1) where ‘m’ is the length of text2.

Therefore, the space complexity of the program is O(n * m) because it depends on the lengths of both input strings text1 and text2.

In summary, the provided program has a time complexity of O(n * m) and a space complexity of O(n * m), where ‘n’ and ‘m’ are the lengths of the input strings text1 and text2, respectively.

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