712. Minimum ASCII Delete Sum for Two Strings

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 <= 1000
s1 and s2 consist of lowercase English letters.

Solutions

Solution 1: Dynamic Programming

Thinking

Equalize two strings by deletions whose cost is the ASCII sum of removed characters. Lengths are \(\le 1000\), so a naive recursion over alignments repeats work.

This is edit distance with delete-only: matching last characters cost nothing; otherwise we must drop one side's last character and add its code. An empty prefix costs the whole other string.

Let \(f[i][j]\) be the min cost for the two prefixes. Fill by that transition; the answer is \(f[m][n]\). Time and space \(O(mn)\).

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 \(O(m \times n)\), and the space complexity is \(O(m \times n)\). Where \(m\) and \(n\) are the lengths of \(s_1\) and \(s_2\) respectively.

Similar problems:

1143. Longest Common Subsequence

Python3JavaC++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]

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];
    }
}

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];
    }
};

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]
}

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];
}

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]
    }
}

/**
 * @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];
};

712. Minimum ASCII Delete Sum for Two Strings

Description

Solutions

Solution 1: Dynamic Programming

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