Medium
,String
,DP
1143. 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.
"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:
text1.length
, text2.length
<= 1000text1
and text2
consist of only lowercase English characters.
function longestCommonSubsequence(text1, text2) {
const m = text1.length;
const n = text2.length;
const dp = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
for (let i = 1; i <= m; i++) {
for (let j = 1; j <= n; j++) {
if (text1[i - 1] === text2[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[m][n];
}
Marsgoat Dec 15, 2022
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
m, n = len(text1), len(text2)
dp = [[0] * (n+1) for _ in range(m+1)]
for i in range(1, len(text1)+1):
for j in range(1, len(text2)+1):
if text1[i-1] == text2[j-1]:
dp[i][j] = dp[i-1][j-1] + 1
else:
dp[i][j] = max(dp[i-1][j], dp[i][j-1])
return dp[m][n]
Kobe Dec 15, 2022