String
Dynamic Programming
Description
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:
Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')
Example 2:
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.
Solutions
Solution 1: Dynamic Programming
We define $f[i][j]$ as the minimum number of operations to convert $word1$ of length $i$ to $word2$ of length $j$. $f[i][0] = i$, $f[0][j] = j$, $i \in [1, m], j \in [0, n]$.
We consider $f[i][j]$:
If $word1[i - 1] = word2[j - 1]$, then we only need to consider the minimum number of operations to convert $word1$ of length $i - 1$ to $word2$ of length $j - 1$, so $f[i][j] = f[i - 1][j - 1]$;
Otherwise, we can consider insert, delete, and replace operations, then $f[i][j] = \min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1$.
Finally, we can get the state transition equation:
$$
f[i][j] = \begin{cases}
i, & \textit{if } j = 0 \
j, & \textit{if } i = 0 \
f[i - 1][j - 1], & \textit{if } word1[i - 1] = word2[j - 1] \
\min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1, & \textit{otherwise}
\end{cases}
$$
Finally, we return $f[m][n]$.
The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. $m$ and $n$ are the lengths of $word1$ and $word2$ respectively.
Python3 Java C++ Go TypeScript JavaScript
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14 class Solution :
def minDistance ( self , word1 : str , word2 : str ) -> int :
m , n = len ( word1 ), len ( word2 )
f = [[ 0 ] * ( n + 1 ) for _ in range ( m + 1 )]
for j in range ( 1 , n + 1 ):
f [ 0 ][ j ] = j
for i , a in enumerate ( word1 , 1 ):
f [ i ][ 0 ] = i
for j , b in enumerate ( word2 , 1 ):
if a == b :
f [ i ][ j ] = f [ i - 1 ][ j - 1 ]
else :
f [ i ][ j ] = min ( f [ i - 1 ][ j ], f [ i ][ j - 1 ], f [ i - 1 ][ j - 1 ]) + 1
return f [ m ][ n ]
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20 class Solution {
public int minDistance ( String word1 , String word2 ) {
int m = word1 . length (), n = word2 . length ();
int [][] f = new int [ m + 1 ][ n + 1 ] ;
for ( int j = 1 ; j <= n ; ++ j ) {
f [ 0 ][ j ] = j ;
}
for ( int i = 1 ; i <= m ; ++ i ) {
f [ i ][ 0 ] = i ;
for ( int j = 1 ; j <= n ; ++ j ) {
if ( word1 . charAt ( i - 1 ) == word2 . charAt ( j - 1 )) {
f [ i ][ j ] = f [ i - 1 ][ j - 1 ] ;
} else {
f [ i ][ j ] = Math . min ( f [ i - 1 ][ j ] , Math . min ( f [ i ][ j - 1 ] , f [ i - 1 ][ j - 1 ] )) + 1 ;
}
}
}
return f [ m ][ n ] ;
}
}
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21 class Solution {
public :
int minDistance ( string word1 , string word2 ) {
int m = word1 . size (), n = word2 . size ();
int f [ m + 1 ][ n + 1 ];
for ( int j = 0 ; j <= n ; ++ j ) {
f [ 0 ][ j ] = j ;
}
for ( int i = 1 ; i <= m ; ++ i ) {
f [ i ][ 0 ] = i ;
for ( int j = 1 ; j <= n ; ++ j ) {
if ( word1 [ i - 1 ] == word2 [ j - 1 ]) {
f [ i ][ j ] = f [ i - 1 ][ j - 1 ];
} else {
f [ i ][ j ] = min ({ f [ i - 1 ][ j ], f [ i ][ j - 1 ], f [ i - 1 ][ j - 1 ]}) + 1 ;
}
}
}
return f [ m ][ n ];
}
};
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21 func minDistance ( word1 string , word2 string ) int {
m , n := len ( word1 ), len ( word2 )
f := make ([][] int , m + 1 )
for i := range f {
f [ i ] = make ([] int , n + 1 )
}
for j := 1 ; j <= n ; j ++ {
f [ 0 ][ j ] = j
}
for i := 1 ; i <= m ; i ++ {
f [ i ][ 0 ] = i
for j := 1 ; j <= n ; j ++ {
if word1 [ i - 1 ] == word2 [ j - 1 ] {
f [ i ][ j ] = f [ i - 1 ][ j - 1 ]
} else {
f [ i ][ j ] = min ( f [ i - 1 ][ j ], min ( f [ i ][ j - 1 ], f [ i - 1 ][ j - 1 ])) + 1
}
}
}
return f [ m ][ n ]
}
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21 function minDistance ( word1 : string , word2 : string ) : number {
const m = word1 . length ;
const n = word2 . length ;
const f : number [][] = Array ( m + 1 )
. fill ( 0 )
. map (() => Array ( n + 1 ). fill ( 0 ));
for ( let j = 1 ; j <= n ; ++ j ) {
f [ 0 ][ j ] = j ;
}
for ( let i = 1 ; i <= m ; ++ i ) {
f [ i ][ 0 ] = i ;
for ( let j = 1 ; j <= n ; ++ j ) {
if ( word1 [ i - 1 ] === word2 [ j - 1 ]) {
f [ i ][ j ] = f [ i - 1 ][ j - 1 ];
} else {
f [ i ][ j ] = Math . min ( f [ i - 1 ][ j ], f [ i ][ j - 1 ], f [ i - 1 ][ j - 1 ]) + 1 ;
}
}
}
return f [ m ][ n ];
}
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26 /**
* @param {string} word1
* @param {string} word2
* @return {number}
*/
var minDistance = function ( word1 , word2 ) {
const m = word1 . length ;
const n = word2 . length ;
const f = Array ( m + 1 )
. fill ( 0 )
. map (() => Array ( n + 1 ). fill ( 0 ));
for ( let j = 1 ; j <= n ; ++ j ) {
f [ 0 ][ j ] = j ;
}
for ( let i = 1 ; i <= m ; ++ i ) {
f [ i ][ 0 ] = i ;
for ( let j = 1 ; j <= n ; ++ j ) {
if ( word1 [ i - 1 ] === word2 [ j - 1 ]) {
f [ i ][ j ] = f [ i - 1 ][ j - 1 ];
} else {
f [ i ][ j ] = Math . min ( f [ i - 1 ][ j ], f [ i ][ j - 1 ], f [ i - 1 ][ j - 1 ]) + 1 ;
}
}
}
return f [ m ][ n ];
};
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