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72. Edit Distance

Posted on Edited on

https://leetcode.com/problems/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

Example 1:

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Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation: 
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

Example 2:

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Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation: 
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

Constraints:

  • 0 <= word1.length, word2.length <= 500
  • word1 and word2 consist of lowercase English letters

Let P[i][j] denote the cost of transforming a[0..i-1] to b[0..j-1].
For current operation, we have several choices.

Assume we have solved P[i-1][j-1], P[i-1][j], P[i][j-1]. Now we try to solve P[i][j].
if current operation is delete, then we must have transformed a[0..i-2] to b[0..j-1]
if current operation is insert, then we must have transformed a[0..i-1] to b[0..j-2]
Otherwise, then we have transformed a[0..i-1] to b[0..j-1].
If a[i-1]=b[j-1], then current operation is do nothing.
If a[i-1]!=b[j-1], then current operation is replacement.

Thus for iāˆˆ[1,n], jāˆˆ[1,m], we have
$$
P[i][j]=\begin{cases}
P[i-1][j]+1
& \
P[i][j-1]+1
& \
P[i-1][j-1]
& a[i-1]=b[j-1] \
P[i-1][j-1]+1
& a[i-1]!=b[j-1]
\end{cases}
$$

Base cases:

The cost of transforming a[0..i-1] to empty string is always doing deletion i times .

Thus P[i][0]=i

The cost of transforming emtpy string to b[0..j-1] is always doing insertion j times.

Thus P[0][j]=j

Order of Computation and Final Solution

Before we solve P[i][j], we need to solve P[i-1][j],P[i][j-1],P[i][j], thus the order should be increasing order of i and j.

The final solution is P[n][m]

Time Complexity

There are n*m subproblems, thus the time complexity is O(nm).

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class Solution:
    def minDistance(self, a: str, b: str) -> int:
        n=len(a)
        m=len(b)
        P=[[0]*(m+1) for _ in range(n+1)]
        if m==0:
            return n
        if n==0:
            return m
                
        for i in range(1,n+1):
            P[i][0]=i
        for j in range(1,m+1):
            P[0][j]=j
       
        
        for i in range(1,n+1):
            for j in range(1,m+1):
                if a[i-1]==b[j-1]:
                    P[i][j]=min(
                        P[i-1][j]+1, #delete
                        P[i][j-1]+1, #insert
                        P[i-1][j-1]  
                    )
                else:
                    P[i][j]=min(
                        P[i-1][j]+1,
                        P[i][j-1]+1,
                        P[i-1][j-1]+1 #replace
                    )
        return P[n][m]