56 lines
2.0 KiB
C++
56 lines
2.0 KiB
C++
// Source : https://leetcode.com/problems/minimum-ascii-delete-sum-for-two-strings/
|
|
// Author : Hao Chen
|
|
// Date : 2019-01-30
|
|
|
|
/*****************************************************************************************************
|
|
*
|
|
* 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 .
|
|
* All elements of each string will have an ASCII value in [97, 122].
|
|
******************************************************************************************************/
|
|
|
|
class Solution {
|
|
public:
|
|
int minimumDeleteSum(string s1, string s2) {
|
|
// Dynamic Programm - simlar with : Edit Distance
|
|
vector < vector <int> > dp ( s1.size()+1, vector<int>( s2.size()+1, 0) );
|
|
|
|
// s1 is row, s2 is column
|
|
for (int i=1; i<=s2.size(); i++) dp[0][i] = dp[0][i-1] + s2[i-1];
|
|
for (int i=1; i<=s1.size(); i++) dp[i][0] = dp[i-1][0] + s1[i-1];
|
|
|
|
for (int i=1; i<=s1.size(); i++){
|
|
for (int j=1; j<=s2.size(); 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[s1.size()][s2.size()];
|
|
|
|
}
|
|
};
|