Given two strings s1, s2, find 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.
Note:
- 0 < s1.length, s2.length <= 1000.
- All elements of each string will have an ASCII value in [97, 122].
This is clearly a DP problem.
dp[i][j] is the cost for s1.substr(0,i) and s2.substr(0, j). Note s1[i], s2[j] not included in the substring.
Base case: dp[0][0] = 0
target: dp[m][n]
if s1[i-1] = s2[j-1] // no deletion
dp[i][j] = dp[i-1][j-1];
else // delete either s1[i-1] or s2[j-1]
dp[i][j] = min(dp[i-1][j]+s1[i-1], dp[i][j-1]+s2[j-1]);
We can use a 2D vector, or an optimized O(n) extra space. See below. The run time is O(mn).
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, 0));
for (int j = 1; j <= n; j++)
dp[0][j] = dp[0][j-1]+s2[j-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++) {
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];
}
};
Optimized O(n) extra space
class Solution {
public:
int minimumDeleteSum(string s1, string s2) {
int m = s1.size(), n = s2.size();
vector<int> dp(n+1, 0);
for (int j = 1; j <= n; j++)
dp[j] = dp[j-1]+s2[j-1];
for (int i = 1; i <= m; i++) {
int t1 = dp[0];
dp[0] += s1[i-1];
for (int j = 1; j <= n; j++) {
int t2 = dp[j];
dp[j] = s1[i-1] == s2[j-1]? t1:min(dp[j]+s1[i-1], dp[j-1]+s2[j-1]);
t1 = t2;
}
}
return dp[n];
}
};