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

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

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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);
}