Given two strings
text2, return the length of their longest common subsequence.
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. (eg, “ace” is a subsequence of “abcde” while “aec” is not). A common subsequence of two strings is a subsequence that is common to both strings.
If there is no common subsequence, return 0.
Input: text1 = "abcde", text2 = "ace" Output: 3 Explanation: The longest common subsequence is "ace" and its length is 3.
Input: text1 = "abc", text2 = "abc" Output: 3 Explanation: The longest common subsequence is "abc" and its length is 3.
Input: text1 = "abc", text2 = "def" Output: 0 Explanation: There is no such common subsequence, so the result is 0.
1 <= text1.length <= 1000
1 <= text2.length <= 1000
- The input strings consist of lowercase English characters only.
Try dynamic programming. DP[i][j] represents the longest common subsequence of text1[0 … i] & text2[0 … j].
DP[i][j] = DP[i – 1][j – 1] + 1 , if text1[i] == text2[j] DP[i][j] = max(DP[i – 1][j], DP[i][j – 1]) , otherwise
class Solution: def longestCommonSubsequence(self, text1: str, text2: str) -> int: f = [[0 for j in range(0, len(text2) + 1)] for i in range(0, len(text1) + 1)] for i in range(1, len(text1) + 1): for j in range (1, len(text2) + 1): if text1[i - 1] == text2[j - 1]: f[i][j] = f[i - 1][j - 1] + 1 else: f[i][j] = max(f[i - 1][j], f[i][j - 1]) return f[len(text1)][len(text2)]
- Time complexity: O(MN). We’re solving M * Nsubproblems. Solving each subproblem is an O(1)operation.
- Space complexity: O(MN).We’re allocating a 2D array of size M * N to save the answers to subproblems.