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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: 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'.
Example 4:
Input: s1 = "112s", s2 = "g841" Output: true Explanation: 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: false Explanation: 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
ands2
consist of digits1-9
(inclusive), and lowercase English letters only.- The number of consecutive digits in
s1
ands2
does not exceed3
.
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);
}
};