Description
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.
Solutions
Solution 1: Dynamic Programming
We define f[i][j] as the minimum sum of ASCII values of deleted characters required to make the first i characters of s_1 equal to the first j characters of s_2. The answer is f[m][n].
If s_1[i-1] = s_2[j-1], then f[i][j] = f[i-1][j-1]. Otherwise, we can delete either s_1[i-1] or s_2[j-1] to minimize f[i][j]. Therefore, the state transition equation is as follows:
$$
f[i][j]=
\begin{cases}
f[i-1][j-1], & s_1[i-1] = s_2[j-1] \
min(f[i-1][j] + s_1[i-1], f[i][j-1] + s_2[j-1]), & s_1[i-1] \neq s_2[j-1]
\end{cases}
$$
The initial state is f[0][j] = f[0][j-1] + s_2[j-1], f[i][0] = f[i-1][0] + s_1[i-1].
Finally, return f[m][n].
The time complexity is O(m × n), and the space complexity is O(m × n). Where m and n are the lengths of s_1 and s_2 respectively.
Similar problems:
PythonJavaC++GoTypeScriptRustJavaScript
class Solution:
def minimumDeleteSum(self, s1: str, s2: str) -> int:
m, n = len(s1), len(s2)
f = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
f[i][0] = f[i - 1][0] + ord(s1[i - 1])
for j in range(1, n + 1):
f[0][j] = f[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]:
f[i][j] = f[i - 1][j - 1]
else:
f[i][j] = min(
f[i - 1][j] + ord(s1[i - 1]), f[i][j - 1] + ord(s2[j - 1])
)
return f[m][n](code-box)
class Solution {
public int minimumDeleteSum(String s1, String s2) {
int m = s1.length(), n = s2.length();
int[][] f = new int[m + 1][n + 1];
for (int i = 1; i <= m; ++i) {
f[i][0] = f[i - 1][0] + s1.charAt(i - 1);
}
for (int j = 1; j <= n; ++j) {
f[0][j] = f[0][j - 1] + s2.charAt(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)) {
f[i][j] = f[i - 1][j - 1];
} else {
f[i][j]
= Math.min(f[i - 1][j] + s1.charAt(i - 1), f[i][j - 1] + s2.charAt(j - 1));
}
}
}
return f[m][n];
}
}(code-box)
class Solution {
public:
int minimumDeleteSum(string s1, string s2) {
int m = s1.size(), n = s2.size();
int f[m + 1][n + 1];
memset(f, 0, sizeof f);
for (int i = 1; i <= m; ++i) {
f[i][0] = f[i - 1][0] + s1[i - 1];
}
for (int j = 1; j <= n; ++j) {
f[0][j] = f[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]) {
f[i][j] = f[i - 1][j - 1];
} else {
f[i][j] = min(f[i - 1][j] + s1[i - 1], f[i][j - 1] + s2[j - 1]);
}
}
}
return f[m][n];
}
};(code-box)
func minimumDeleteSum(s1 string, s2 string) int {
m, n := len(s1), len(s2)
f := make([][]int, m+1)
for i := range f {
f[i] = make([]int, n+1)
}
for i, c := range s1 {
f[i+1][0] = f[i][0] + int(c)
}
for j, c := range s2 {
f[0][j+1] = f[0][j] + int(c)
}
for i := 1; i <= m; i++ {
for j := 1; j <= n; j++ {
if s1[i-1] == s2[j-1] {
f[i][j] = f[i-1][j-1]
} else {
f[i][j] = min(f[i-1][j]+int(s1[i-1]), f[i][j-1]+int(s2[j-1]))
}
}
}
return f[m][n]
}(code-box)
function minimumDeleteSum(s1: string, s2: string): number {
const m = s1.length;
const n = s2.length;
const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
for (let i = 1; i <= m; ++i) {
f[i][0] = f[i - 1][0] + s1[i - 1].charCodeAt(0);
}
for (let j = 1; j <= n; ++j) {
f[0][j] = f[0][j - 1] + s2[j - 1].charCodeAt(0);
}
for (let i = 1; i <= m; ++i) {
for (let j = 1; j <= n; ++j) {
if (s1[i - 1] === s2[j - 1]) {
f[i][j] = f[i - 1][j - 1];
} else {
f[i][j] = Math.min(
f[i - 1][j] + s1[i - 1].charCodeAt(0),
f[i][j - 1] + s2[j - 1].charCodeAt(0),
);
}
}
}
return f[m][n];
}(code-box)
impl Solution {
pub fn minimum_delete_sum(s1: String, s2: String) -> i32 {
let m: usize = s1.len();
let n: usize = s2.len();
let b1 = s1.as_bytes();
let b2 = s2.as_bytes();
let mut f: Vec<Vec<i32>> = vec![vec![0; n + 1]; m + 1];
for i in 1..=m {
f[i][0] = f[i - 1][0] + b1[i - 1] as i32;
}
for j in 1..=n {
f[0][j] = f[0][j - 1] + b2[j - 1] as i32;
}
for i in 1..=m {
for j in 1..=n {
if b1[i - 1] == b2[j - 1] {
f[i][j] = f[i - 1][j - 1];
} else {
f[i][j] = std::cmp::min(
f[i - 1][j] + b1[i - 1] as i32,
f[i][j - 1] + b2[j - 1] as i32,
);
}
}
}
f[m][n]
}
}(code-box)
/**
* @param {string} s1
* @param {string} s2
* @return {number}
*/
var minimumDeleteSum = function (s1, s2) {
const m = s1.length;
const n = s2.length;
const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
for (let i = 1; i <= m; ++i) {
f[i][0] = f[i - 1][0] + s1[i - 1].charCodeAt(0);
}
for (let j = 1; j <= n; ++j) {
f[0][j] = f[0][j - 1] + s2[j - 1].charCodeAt(0);
}
for (let i = 1; i <= m; ++i) {
for (let j = 1; j <= n; ++j) {
if (s1[i - 1] === s2[j - 1]) {
f[i][j] = f[i - 1][j - 1];
} else {
f[i][j] = Math.min(
f[i - 1][j] + s1[i - 1].charCodeAt(0),
f[i][j - 1] + s2[j - 1].charCodeAt(0),
);
}
}
}
return f[m][n];
};(code-box)
