Given two strings s1
and s2
, return the lowest ASCII sum of deleted characters to make two strings equal.
Example 1:
Input: s1 = "sea", s2 = "eat" Output: 231 Explanation: Deleting "s" from "sea" adds the ASCII value of "s" (115) to the sum. Deleting "t" from "eat" adds 116 to the sum. At the end, both strings are equal, and 115 + 116 = 231 is the minimum sum possible to achieve this.
Example 2:
Input: s1 = "delete", s2 = "leet" Output: 403 Explanation: Deleting "dee" from "delete" to turn the string into "let", adds 100[d] + 101[e] + 101[e] to the sum. Deleting "e" from "leet" adds 101[e] to the sum. At the end, both strings are equal to "let", and the answer is 100+101+101+101 = 403. If instead we turned both strings into "lee" or "eet", we would get answers of 433 or 417, which are higher.
Constraints:
1 <= s1.length, s2.length <= 1000
s1
ands2
consist of lowercase English letters.
Dynamic programming.
class Solution:
def minimumDeleteSum(self, s1: str, s2: str) -> int:
m, n = len(s1), len(s2)
dp = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
dp[i][0] = dp[i - 1][0] + ord(s1[i - 1])
for j in range(1, n + 1):
dp[0][j] = dp[0][j - 1] + ord(s2[j - 1])
for i in range(1, m + 1):
for j in range(1, n + 1):
if s1[i - 1] == s2[j - 1]:
dp[i][j] = dp[i - 1][j - 1]
else:
dp[i][j] = min(dp[i - 1][j] + ord(s1[i - 1]),
dp[i][j - 1] + ord(s2[j - 1]))
return dp[-1][-1]
class Solution {
public int minimumDeleteSum(String s1, String s2) {
int m = s1.length(), n = s2.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 1; i <= m; ++i) {
dp[i][0] = dp[i - 1][0] + s1.codePointAt(i - 1);
}
for (int j = 1; j <= n; ++j) {
dp[0][j] = dp[0][j - 1] + s2.codePointAt(j - 1);
}
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = Math.min(dp[i - 1][j] + s1.codePointAt(i - 1), dp[i][j - 1] + s2.codePointAt(j - 1));
}
}
}
return dp[m][n];
}
}
class Solution {
public:
int minimumDeleteSum(string s1, string s2) {
int m = s1.size(), n = s2.size();
vector<vector<int>> dp(m + 1, vector<int>(n + 1));
for (int i = 1; i <= m; ++i) dp[i][0] = dp[i - 1][0] + s1[i - 1];
for (int j = 1; j <= n; ++j) dp[0][j] = dp[0][j - 1] + s2[j - 1];
for (int i = 1; i <= m; ++i)
{
for (int j = 1; j <= n; ++j)
{
if (s1[i - 1] == s2[j - 1]) dp[i][j] = dp[i - 1][j - 1];
else dp[i][j] = min(dp[i - 1][j] + s1[i - 1], dp[i][j - 1] + s2[j - 1]);
}
}
return dp[m][n];
}
};