583. Delete Operation for Two Strings83
1. 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.
2. Example
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
3. Constraints
- 1 <= word1.length, word2.length <= 500
- word1 and word2 consist of only lowercase English letters.
4. Solutions
Dynamic Programming
m = word1.size(), n = word2.size()
Time complexity: O(mn)
Space complexity: O(min(m, n))
class Solution {
public:
int minDistance(const string &word1, const string &word2) {
const string &short_word = word1.size() > word2.size() ? word2 : word1;
const string &long_word = word1.size() > word2.size() ? word1 : word2;
array<vector<int>, 2> dp = {
vector<int>(short_word.size() + 1), vector<int>(short_word.size() + 1)};
// dp[i][j] means LCS length of word1[0, i] and word1[0, j]
for (int i = 1; i <= long_word.size(); ++i) {
for (int j = 1; j <= short_word.size(); ++j) {
dp[1][j] = long_word[i - 1] == short_word[j - 1] ? dp[0][j - 1] + 1
: max(dp[0][j], dp[1][j - 1]);
}
swap(dp[0], dp[1]); // avoid copy
}
return short_word.size() + long_word.size() - 2 * dp[0].back();
}
};