583. Delete Operation for Two Strings 72. Edit Distance

583. Delete Operation for Two Strings

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.

1. dpij: the minimum number of times to deleted a letter。

the string word1 that ends in i-1

the string word2 that ends in j-1

2. recursive formula

When word1i - 1 and word2j - 1 are the same

When word1i - 1 and word2j - 1 are not the same

When word1i - 1 is the same as word2j - 1, dpij = dpi - 1j - 1;

When word1i - 1 is not the same as word2j - 1, there are three cases:

Case 1: delete word1i - 1, the minimum number of operations is dpi - 1j + 1

Case 2: delete word2j - 1, the minimum number of operations is dpij - 1 + 1

**Case 3:**delete word1i - 1 and word2j - 1 at the same time, the minimum number of operations for dpi - 1j - 1 + 2

then finally of course the minimum value, so when word1i - 1 and word2j - 1 are not the same, recursive formula: dpij = min({dpi - 1j - 1 + 2, dpi - 1j + 1, dpij - 1 + 1});

Since dpij - 1 + 1 = dpi - 1j - 1 + 2, the recursive formula can be simplified to

dpij = min(dpi - 1j, dpij - 1) + 1;

DP:

Time complexity: O(m x n)

Space complexity: O(m x n)

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m = len(word1)
        n = len(word2)

        dp = [[0] * (n + 1) for _ in range(m + 1)]

        for i in range(m + 1): # m+1 不是 m
            dp[i][0] += i
        for j in range(n + 1):
            dp[0][j] += j
        
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if word1[i - 1] == word2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1]
                else:
                    dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1#dp[i][j] = min(dp[i-1][j-1] + 2, dp[i-1][j] + 1, dp[i][j-1] + 1)
        
        return dp[-1][-1]

72. Edit Distance

Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.

You have the following three operations permitted on a word:

• Insert a character
• Delete a character
• Replace a character

recursive formula:

if (word1i - 1 == word2j - 1)

不操作 dpij = dpi - 1j - 1

if (word1i - 1 != word2j - 1)

增 dpij = dpi - 1j + 1,dpij = dpij - 1 + 1 Adding an element to word2 is equivalent to removing an element from word1

删 dpij = dpi - 1j + 1,dpij = dpij - 1 + 1

换 dpij = dpi - 1j - 1 + 1

DP:

Time complexity: O(m x n)

Space complexity: O(m x n)

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m = len(word1)
        n = len(word2)

        dp = [[0] * (n + 1) for _ in range(m + 1)]

        for i in range(m + 1):
            dp[i][0] = i
        for j in range(n + 1):
            dp[0][j] = j
        
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if word1[i - 1] == word2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1]
                else:
                    dp[i][j] = min(dp[i - 1][j - 1] + 1, dp[i - 1][j] + 1, dp[i][j - 1] + 1)
        
        return dp[-1][-1]