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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 nonempty 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 19
(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
ands2
consist of digits19
(inclusive), and lowercase English letters only. The number of consecutive digits in
s1
ands2
does not exceed3
.
Solutions
Dynamic Programming

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); }