[LeetCode] 583. Delete Operation for Two Strings

 轉自LeetCode

Given 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 <= 500
• word1 and word2 consist of only lowercase English letters.

Solution

需要用到 1143. Longest Common Subsequence 的想法

如果能找到兩個字串的 longest common subsequence(LCS)

那麼最後的答案就是 word1.length + word2.length - LCS.length * 2

kotlin

	class Solution {
	fun minDistance(word1: String, word2: String): Int {
	// find longest common subsequence
	val dp = Array(word1.length+1) { IntArray(word2.length+1) {0} }
	for(i in 1..word1.length) {
	for(j in 1..word2.length) {
	if(word1[i-1] == word2[j-1]) {
	dp[i][j] = dp[i-1][j-1] + 1
	} else {
	dp[i][j] = Math.max(dp[i][j-1], dp[i-1][j])
	}
	}
	}
	return word1.length + word2.length - dp.last().last() * 2
	}
	}

view raw delete_operation_for_two_strings.kt hosted with ❤ by GitHub