Description
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 <= 500word1 and word2 consist of only lowercase English letters.
Solutions
Solution 1: Dynamic Programming
We define f[i][j] as the minimum number of deletions required to make the first i characters of the string word1 and the first j characters of the string word2 the same. The answer is f[m][n], where m and n are the lengths of the strings word1 and word2, respectively.
Initially, if j = 0, then f[i][0] = i; if i = 0, then f[0][j] = j.
When i, j > 0, if word1[i - 1] = word2[j - 1], then f[i][j] = f[i - 1][j - 1]; otherwise, f[i][j] = min(f[i - 1][j], f[i][j - 1]) + 1.
Finally, return f[m][n].
The time complexity is O(m × n), and the space complexity is O(m × n). Here, m and n are the lengths of the strings word1 and word2, respectively.
PythonJavaC++GoTypeScriptRust
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 i in range(1, m + 1):
f[i][0] = i
for j in range(1, n + 1):
f[0][j] = j
for i, a in enumerate(word1, 1):
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]) + 1
return f[m][n](code-box)
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 i = 0; i <= m; ++i) {
f[i][0] = i;
}
for (int j = 0; j <= n; ++j) {
f[0][j] = j;
}
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
char a = word1.charAt(i - 1);
char b = word2.charAt(j - 1);
if (a == b) {
f[i][j] = f[i - 1][j - 1];
} else {
f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + 1;
}
}
}
return f[m][n];
}
}(code-box)
class Solution {
public:
int minDistance(string word1, string word2) {
int m = word1.length(), n = word2.length();
vector<vector<int>> f(m + 1, vector<int>(n + 1));
for (int i = 0; i <= m; ++i) {
f[i][0] = i;
}
for (int j = 0; j <= n; ++j) {
f[0][j] = j;
}
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
char a = word1[i - 1];
char b = word2[j - 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]) + 1;
}
}
}
return f[m][n];
}
};(code-box)
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)
f[i][0] = i
}
for j := 1; j <= n; j++ {
f[0][j] = j
}
for i := 1; i <= m; i++ {
for j := 1; j <= n; j++ {
a, b := word1[i-1], word2[j-1]
if a == b {
f[i][j] = f[i-1][j-1]
} else {
f[i][j] = 1 + min(f[i-1][j], f[i][j-1])
}
}
}
return f[m][n]
}(code-box)
function minDistance(word1: string, word2: string): number {
const m = word1.length;
const n = word2.length;
const f: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
for (let i = 1; i <= m; ++i) {
f[i][0] = i;
}
for (let j = 1; j <= n; ++j) {
f[0][j] = j;
}
for (let i = 1; i <= m; ++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]) + 1;
}
}
}
return f[m][n];
}(code-box)
impl Solution {
pub fn min_distance(word1: String, word2: String) -> i32 {
let m = word1.len();
let n = word2.len();
let s: Vec<char> = word1.chars().collect();
let t: Vec<char> = word2.chars().collect();
let mut f = vec![vec![0; n + 1]; m + 1];
for i in 0..=m {
f[i][0] = i as i32;
}
for j in 0..=n {
f[0][j] = j as i32;
}
for i in 1..=m {
for j in 1..=n {
let a = s[i - 1];
let b = t[j - 1];
if a == b {
f[i][j] = f[i - 1][j - 1];
} else {
f[i][j] = std::cmp::min(f[i - 1][j], f[i][j - 1]) + 1;
}
}
}
f[m][n]
}
}(code-box)
