LeetCode 솔루션 분류
[6/14] 583. Delete Operation for Two Strings
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Medium
319749Add to ListShareGiven two strings word1 and word2, return the minimum number of steps required to make word1 and word2 the same.
In one step, you can delete exactly one character in either string.
Example 1:
Input: word1 = "sea", word2 = "eat" Output: 2 Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".
Example 2:
Input: word1 = "leetcode", word2 = "etco" Output: 4
Constraints:
1 <= word1.length, word2.length <= 500word1andword2consist of only lowercase English letters.
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학부유학생님의 댓글
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Runtime: 392 ms, faster than 51.22% of Python3 online submissions for Delete Operation for Two Strings.
Memory Usage: 13.9 MB, less than 92.06% of Python3 online submissions for Delete Operation for Two Strings.
Memory Usage: 13.9 MB, less than 92.06% of Python3 online submissions for Delete Operation for Two Strings.
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
if len(word1) < len(word2):
word1, word2 = word2, word1
long, short = len(word1), len(word2)
prev = [i for i in range(long+1)]
for idx, char in enumerate(word2):
curr = [0]*(long+1)
curr[0] = idx + 1
for i in range(1,long+1):
if char == word1[i-1]:
curr[i] = prev[i-1]
else:
curr[i] = 1+min(prev[i], curr[i-1])
prev = curr
return curr[-1]Coffee님의 댓글
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class Solution {
public int minDistance(String word1, String word2) {
int[][] dp = new int[word1.length() +1][word2.length()+1];
for(int i=0; i<=word1.length(); i++){
for(int j=0; j<=word2.length(); j++){
if(i == 0 || j == 0){
continue;
}
if(word1.charAt(i-1) == word2.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 word1.length() + word2.length() - 2 * dp[word1.length()][word2.length()];
}
}
