README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Hard	https://github.com/doocs/leetcode/edit/main/solution/2000-2099/2060.Check%20if%20an%20Original%20String%20Exists%20Given%20Two%20Encoded%20Strings/README_EN.md	2803	Weekly Contest 265 Q4	String Dynamic Programming

2060. Check if an Original String Exists Given Two Encoded Strings

中文文档

Description

An original string, consisting of lowercase English letters, can be encoded by the following steps:

• Arbitrarily split it into a sequence of some number of non-empty substrings.
• Arbitrarily choose some elements (possibly none) of the sequence, and replace each with its length (as a numeric string).
• Concatenate the sequence as the encoded string.

For example, one way to encode an original string "abcdefghijklmnop" might be:

• Split it as a sequence: ["ab", "cdefghijklmn", "o", "p"].
• Choose the second and third elements to be replaced by their lengths, respectively. The sequence becomes ["ab", "12", "1", "p"].
• Concatenate the elements of the sequence to get the encoded string: "ab121p".

Given two encoded strings s1 and s2, consisting of lowercase English letters and digits 1-9 (inclusive), return true if there exists an original string that could be encoded as both s1 and s2. Otherwise, return false.

Note: The test cases are generated such that the number of consecutive digits in s1 and s2 does not exceed 3.

 

Example 1:

Input: s1 = "internationalization", s2 = "i18n"
Output: true
Explanation: It is possible that "internationalization" was the original string.
- "internationalization" 
  -> Split:       ["internationalization"]
  -> Do not replace any element
  -> Concatenate:  "internationalization", which is s1.
- "internationalization"
  -> Split:       ["i", "nternationalizatio", "n"]
  -> Replace:     ["i", "18",                 "n"]
  -> Concatenate:  "i18n", which is s2

Example 2:

Input: s1 = "l123e", s2 = "44"
Output: true
Explanation: It is possible that "leetcode" was the original string.
- "leetcode" 
  -> Split:      ["l", "e", "et", "cod", "e"]
  -> Replace:    ["l", "1", "2",  "3",   "e"]
  -> Concatenate: "l123e", which is s1.
- "leetcode" 
  -> Split:      ["leet", "code"]
  -> Replace:    ["4",    "4"]
  -> Concatenate: "44", which is s2.

Example 3:

Input: s1 = "a5b", s2 = "c5b"
Output: false
Explanation: It is impossible.
- The original string encoded as s1 must start with the letter 'a'.
- The original string encoded as s2 must start with the letter 'c'.

 

Constraints:

• 1 <= s1.length, s2.length <= 40
• s1 and s2 consist of digits 1-9 (inclusive), and lowercase English letters only.
• The number of consecutive digits in s1 and s2 does not exceed 3.

Solutions

Solution 1

TypeScript

function possiblyEquals(s1: string, s2: string): boolean {
    const n = s1.length,
        m = s2.length;
    let dp: Array<Array<Set<number>>> = Array.from({ length: n + 1 }, v =>
        Array.from({ length: m + 1 }, w => new Set()),
    );
    dp[0][0].add(0);

    for (let i = 0; i <= n; i++) {
        for (let j = 0; j <= m; j++) {
            for (let delta of dp[i][j]) {
                // s1为数字
                let num = 0;
                if (delta <= 0) {
                    for (let p = i; i < Math.min(i + 3, n); p++) {
                        if (isDigit(s1[p])) {
                            num = num * 10 + Number(s1[p]);
                            dp[p + 1][j].add(delta + num);
                        } else {
                            break;
                        }
                    }
                }

                // s2为数字
                num = 0;
                if (delta >= 0) {
                    for (let q = j; q < Math.min(j + 3, m); q++) {
                        if (isDigit(s2[q])) {
                            num = num * 10 + Number(s2[q]);
                            dp[i][q + 1].add(delta - num);
                        } else {
                            break;
                        }
                    }
                }

                // 数字匹配s1为字母
                if (i < n && delta < 0 && !isDigit(s1[i])) {
                    dp[i + 1][j].add(delta + 1);
                }

                // 数字匹配s2为字母
                if (j < m && delta > 0 && !isDigit(s2[j])) {
                    dp[i][j + 1].add(delta - 1);
                }

                // 两个字母匹配
                if (i < n && j < m && delta == 0 && s1[i] == s2[j]) {
                    dp[i + 1][j + 1].add(0);
                }
            }
        }
    }
    return dp[n][m].has(0);
}

function isDigit(char: string): boolean {
    return /^\d{1}$/g.test(char);
}