Description#
You are given two strings s and sub. You are also given a 2D character array mappings where mappings[i] = [oldi, newi] indicates that you may perform the following operation any number of times:
- Replace a character
oldi of sub with newi.
Each character in sub cannot be replaced more than once.
Return true if it is possible to make sub a substring of s by replacing zero or more characters according to mappings. Otherwise, return false.
A substring is a contiguous non-empty sequence of characters within a string.
Example 1:
Input: s = "fool3e7bar", sub = "leet", mappings = [["e","3"],["t","7"],["t","8"]]
Output: true
Explanation: Replace the first 'e' in sub with '3' and 't' in sub with '7'.
Now sub = "l3e7" is a substring of s, so we return true.
Example 2:
Input: s = "fooleetbar", sub = "f00l", mappings = [["o","0"]]
Output: false
Explanation: The string "f00l" is not a substring of s and no replacements can be made.
Note that we cannot replace '0' with 'o'.
Example 3:
Input: s = "Fool33tbaR", sub = "leetd", mappings = [["e","3"],["t","7"],["t","8"],["d","b"],["p","b"]]
Output: true
Explanation: Replace the first and second 'e' in sub with '3' and 'd' in sub with 'b'.
Now sub = "l33tb" is a substring of s, so we return true.
Constraints:
1 <= sub.length <= s.length <= 50000 <= mappings.length <= 1000mappings[i].length == 2oldi != newis and sub consist of uppercase and lowercase English letters and digits.oldi and newi are either uppercase or lowercase English letters or digits.
Solutions#
Solution 1: Hash Table + Enumeration#
First, we use a hash table $d$ to record the set of characters that each character can be replaced with.
Then we enumerate all substrings of length $sub$ in $s$, and judge whether the string $sub$ can be obtained by replacement. If it can, return true, otherwise enumerate the next substring.
At the end of the enumeration, it means that $sub$ cannot be obtained by replacing any substring in $s$, so return false.
The time complexity is $O(m \times n)$, and the space complexity is $O(C^2)$. Here, $m$ and $n$ are the lengths of the strings $s$ and $sub$ respectively, and $C$ is the size of the character set.
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| class Solution:
def matchReplacement(self, s: str, sub: str, mappings: List[List[str]]) -> bool:
d = defaultdict(set)
for a, b in mappings:
d[a].add(b)
for i in range(len(s) - len(sub) + 1):
if all(a == b or a in d[b] for a, b in zip(s[i : i + len(sub)], sub)):
return True
return False
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| class Solution {
public boolean matchReplacement(String s, String sub, char[][] mappings) {
Map<Character, Set<Character>> d = new HashMap<>();
for (var e : mappings) {
d.computeIfAbsent(e[0], k -> new HashSet<>()).add(e[1]);
}
int m = s.length(), n = sub.length();
for (int i = 0; i < m - n + 1; ++i) {
boolean ok = true;
for (int j = 0; j < n && ok; ++j) {
char a = s.charAt(i + j), b = sub.charAt(j);
if (a != b && !d.getOrDefault(b, Collections.emptySet()).contains(a)) {
ok = false;
}
}
if (ok) {
return true;
}
}
return false;
}
}
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| class Solution {
public:
bool matchReplacement(string s, string sub, vector<vector<char>>& mappings) {
unordered_map<char, unordered_set<char>> d;
for (auto& e : mappings) {
d[e[0]].insert(e[1]);
}
int m = s.size(), n = sub.size();
for (int i = 0; i < m - n + 1; ++i) {
bool ok = true;
for (int j = 0; j < n && ok; ++j) {
char a = s[i + j], b = sub[j];
if (a != b && !d[b].count(a)) {
ok = false;
}
}
if (ok) {
return true;
}
}
return false;
}
};
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| func matchReplacement(s string, sub string, mappings [][]byte) bool {
d := map[byte]map[byte]bool{}
for _, e := range mappings {
if d[e[0]] == nil {
d[e[0]] = map[byte]bool{}
}
d[e[0]][e[1]] = true
}
for i := 0; i < len(s)-len(sub)+1; i++ {
ok := true
for j := 0; j < len(sub) && ok; j++ {
a, b := s[i+j], sub[j]
if a != b && !d[b][a] {
ok = false
}
}
if ok {
return true
}
}
return false
}
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Solution 2: Array + Enumeration#
Since the character set only contains uppercase and lowercase English letters and numbers, we can directly use a $128 \times 128$ array $d$ to record the set of characters that each character can be replaced with.
The time complexity is $O(m \times n)$, and the space complexity is $O(C^2)$.
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| class Solution:
def matchReplacement(self, s: str, sub: str, mappings: List[List[str]]) -> bool:
d = [[False] * 128 for _ in range(128)]
for a, b in mappings:
d[ord(a)][ord(b)] = True
for i in range(len(s) - len(sub) + 1):
if all(
a == b or d[ord(b)][ord(a)] for a, b in zip(s[i : i + len(sub)], sub)
):
return True
return False
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| class Solution {
public boolean matchReplacement(String s, String sub, char[][] mappings) {
boolean[][] d = new boolean[128][128];
for (var e : mappings) {
d[e[0]][e[1]] = true;
}
int m = s.length(), n = sub.length();
for (int i = 0; i < m - n + 1; ++i) {
boolean ok = true;
for (int j = 0; j < n && ok; ++j) {
char a = s.charAt(i + j), b = sub.charAt(j);
if (a != b && !d[b][a]) {
ok = false;
}
}
if (ok) {
return true;
}
}
return false;
}
}
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| class Solution {
public:
bool matchReplacement(string s, string sub, vector<vector<char>>& mappings) {
bool d[128][128]{};
for (auto& e : mappings) {
d[e[0]][e[1]] = true;
}
int m = s.size(), n = sub.size();
for (int i = 0; i < m - n + 1; ++i) {
bool ok = true;
for (int j = 0; j < n && ok; ++j) {
char a = s[i + j], b = sub[j];
if (a != b && !d[b][a]) {
ok = false;
}
}
if (ok) {
return true;
}
}
return false;
}
};
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| func matchReplacement(s string, sub string, mappings [][]byte) bool {
d := [128][128]bool{}
for _, e := range mappings {
d[e[0]][e[1]] = true
}
for i := 0; i < len(s)-len(sub)+1; i++ {
ok := true
for j := 0; j < len(sub) && ok; j++ {
a, b := s[i+j], sub[j]
if a != b && !d[b][a] {
ok = false
}
}
if ok {
return true
}
}
return false
}
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