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[12/15] 1143. Longest Common Subsequence

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작성일 2022.12.15 17:15

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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.

Accepted

614.7K

Submissions

<div class="text-label-1 dark:text-dark-label-1 text-sm font-medium" style="border: 0px solid; box-sizing: border-box; --tw-border-spacing-x:0; --tw-border-spacing-y:0; --tw-translate-x:0; --tw-translate-y:0; --tw-rotate:0; --tw-skew-x:0; --tw-skew-y:0; --tw-scale-x:1; --tw-scale-y:1; --tw-pan-x: ; --tw-pan-y: ; --tw-pinch-zoom: ; --tw-scroll-snap-strictness:proximity; --tw-ordinal: ; --tw-slashed-zero: ; --tw-numeric-figure: ; --tw-numeric-spacing: ; --tw-numeric-fraction: ; --tw-ring-inset: ; --tw-ring-offset-width:0px; --tw-ring-offset-color:#fff; --tw-ring-color:rgba(59,130,246,0.5); --tw-ring-offset-shadow:0 0 #0000; --tw-ring-shadow:0 0 #0000; --tw-shadow:0 0 #0000; --tw-shadow-colored:0 0

#Amazon, #Apple, #DoorDash, #Google, #Bloomberg

링크

https://leetcode.com/problems/longest-common-subsequence/ 2591 회 연결

익명
작성일 2022.12.15 17:16

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
    #     a f e
    #   0 0 0 0
    # a 0 1 1 1
    # b 0 1 1 1
    # c 0 1 1 1
    # a 0 1 1 1
    # e 0 1 1 2
        dp = [ [ 0 for _ in range(len(text1)+1) ] for _ in range(len(text2)+1) ]

        for i in range(len(text2)):
            for j in range(len(text1)):
                if text1[j] == text2[i]:
                    dp[i+1][j+1] = dp[i][j]+1
                else:
                    dp[i+1][j+1] = max(dp[i+1][j], dp[i][j+1])
        
        return dp[-1][-1]