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Medium

Longest Common Subsequence

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

Examples

Input:text1 = "abcde", text2 = "ace"
Output:3
The longest common subsequence is "ace" and its length is 3.
Input:text1 = "abc", text2 = "abc"
Output:3
The longest common subsequence is "abc" and its length is 3.
Input:text1 = "abc", text2 = "def"
Output:0
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.

Dynamic Programming (2D Array)

Approach

We use a 2D array `dp` of size `(m+1) x (n+1)` where `dp[i][j]` is the length of the longest common subsequence of `text1[0...i-1]` and `text2[0...j-1]`. We can iterate through both strings. If `text1[i-1] == text2[j-1]`, we add 1 to the result of the LCS without these two characters (`dp[i-1][j-1]`). If they do not match, the LCS is the maximum of the LCS by either skipping the character from `text1` (`dp[i-1][j]`) or skipping the character from `text2` (`dp[i][j-1]`). We initialize the first row and column with 0s.

Complexity Analysis

Time Complexity
O(m * n)
Space Complexity
O(m * n)

Time and space complexity are both O(m * n). The space complexity can be optimized to O(min(m, n)) by keeping only the previous row/column of the DP table.

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