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 <= 1000s1 and s2 consist of lowercase English letters.class Solution {
public:
// Placebo
int minimumDeleteSum(string s1, string s2) {
// Make sure s2 is smaller string
if (s1.length() < s2.length()) {
return minimumDeleteSum(s2, s1);
}
// Case for empty s1
int m = s1.length(), n = s2.length();
vector<int> currRow(n + 1);
for (int j = 1; j <= n; j++) {
currRow[j] = currRow[j - 1] + s2[j - 1];
}
// Compute answer row-by-row
for (int i = 1; i <= m; i++) {
int diag = currRow[0];
currRow[0] += s1[i - 1];
for (int j = 1; j <= n; j++) {
int answer;
// If characters are the same, the answer is top-left-diagonal value
if (s1[i - 1] == s2[j - 1]) {
answer = diag;
}
// Otherwise, the answer is minimum of top and left values with
// deleted character's ASCII value
else {
answer = min(
s1[i - 1] + currRow[j],
s2[j - 1] + currRow[j - 1]
);
}
// Before overwriting currRow[j] with answer, save it in diag
// for the next column
diag = currRow[j];
currRow[j] = answer;
}
}
return currRow[n];
}
};