Given two strings
text2, return the length of their longest common subsequence. If there is no common subsequence, return
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
A common subsequence of two strings is a subsequence that is common to both strings.
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, text2.length <= 1000
text2consist of only lowercase English characters.
Use dynamic programming approach to solve this problem. f[i][j] represents the longest common subsequence between text1[0 : i] and text2[0 : j].
If text1[i] == text2[j], then the longest common subsequence just increase from previous position from both words by 1: f[i][j] = f[i – 1][j – 1] + 1.
Otherwise, if text1[i] != text2[j], the longest common subsequence would be the longest between text[i – 1] and text2[j] or text1[i] and text2[j – 1].
class Solution: def longestCommonSubsequence(self, text1: str, text2: str) -> int: m = len(text1) n = len(text2) f =  for i in range(m + 1): f.append() for j in range(n + 1): f[i].append(0) for i in range(m): for j in range(n): if text1[i] == text2[j]: f[i + 1][j + 1] = f[i][j] + 1 else: f[i + 1][j + 1] = max(f[i][j + 1], f[i + 1][j]) return f[m][n]
- Time complexity: O(MN). We’re solving M * N subproblems. 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.