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Formatted question description: https://leetcode.ca/all/2060.html

# 2060. Check if an Original String Exists Given Two Encoded Strings (Hard)

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:trueExplanation: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:trueExplanation: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:falseExplanation: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'.

**Example 4:**

Input:s1 = "112s", s2 = "g841"Output:trueExplanation:It is possible that "gaaaaaaaaaaaas" was the original string - "gaaaaaaaaaaaas" -> Split: ["g", "aaaaaaaaaaaa", "s"] -> Replace: ["1", "12", "s"] -> Concatenate: "112s", which is s1. - "gaaaaaaaaaaaas" -> Split: ["g", "aaaaaaaa", "aaaa", "s"] -> Replace: ["g", "8", "4", "1"] -> Concatenate: "g841", which is s2.

**Example 5:**

Input:s1 = "ab", s2 = "a2"Output:falseExplanation:It is impossible. - The original string encoded as s1 has two letters. - The original string encoded as s2 has three letters.

**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`

.

**Similar Questions**:

## Solution 1. DP

```
// OJ: https://leetcode.com/problems/check-if-an-original-string-exists-given-two-encoded-strings/
// Time: O(C * MN) where `C` is the range of the possible length differences. It's [-10,000, 10,000] in this problem
// Space: O(C * MN)
// Ref: https://leetcode-cn.com/problems/check-if-an-original-string-exists-given-two-encoded-strings/solution/dong-tai-gui-hua-ji-lu-ke-neng-de-chang-o87gp/
class Solution {
public:
bool possiblyEquals(string s, string t) {
int M = s.size(), N = t.size();
unordered_set<int> dp[41][41] = {};
dp[0][0].insert(0);
for (int i = 0; i <= M; ++i) {
for (int j = 0; j <= N; ++j) {
for (int delta : dp[i][j]) {
int num = 0;
for (int p = i; p < M && isdigit(s[p]); ++p) {
num = num * 10 + (s[p] - '0');
dp[p + 1][j].insert(delta + num);
}
num = 0;
for (int p = j; p < N && isdigit(t[p]); ++p) {
num = num * 10 + (t[p] - '0');
dp[i][p + 1].insert(delta - num);
}
if (i < M && delta < 0 && isalpha(s[i])) dp[i + 1][j].insert(delta + 1);
if (j < N && delta > 0 && isalpha(t[j])) dp[i][j + 1].insert(delta - 1);
if (i < M && j < N && delta == 0 && s[i] == t[j]) dp[i + 1][j + 1].insert(0);
}
}
}
return dp[M][N].count(0);
}
};
```

Or

```
// OJ: https://leetcode.com/problems/check-if-an-original-string-exists-given-two-encoded-strings/
// Time: O(MNC)
// Space: O(MNC) where C is the maximum possible difference of digits. Since there are at most 3 consecutive digits, the difference is at most 1000 (then it must try to match a letter)
// Ref: https://leetcode-cn.com/problems/check-if-an-original-string-exists-given-two-encoded-strings/solution/ji-yi-hua-sou-suo-by-endlesscheng-ll3r/
class Solution {
public:
bool possiblyEquals(string s, string t) {
int M = s.size(), N = t.size();
const int mx = 2001, bias = 1000;
bool seen[41][41][mx] = {};
function <bool(int, int, int)> dfs = [&](int i, int j, int d) {
if (i == M && j == N) return d == 0;
if (i > M || j > N) return false;
if (seen[i][j][d + bias]) return false;
seen[i][j][d + bias] = true;
if (d == 0 && s[i] == t[j] && dfs(i + 1, j + 1, 0)) return true;
if (d <= 0) {
if (isdigit(s[i])) {
for (int p = i, num = 0; p < M && isdigit(s[p]); ++p) {
num = num * 10 + (s[p] - '0');
if (dfs(p + 1, j, d + num)) return true;
}
} else if (d < 0 && dfs(i + 1, j, d + 1)) return true;
}
if (d >= 0) {
if (isdigit(t[j])) {
for (int p = j, num = 0; p < N && isdigit(t[p]); ++p) {
num = num * 10 + (t[p] - '0');
if (dfs(i, p + 1, d - num)) return true;
}
} else if (d > 0 && dfs(i, j + 1, d - 1)) return true;
}
return false;
};
return dfs(0, 0, 0);
}
};
```