2301. Match Substring After Replacement

Description

You are given two strings s and sub. You are also given a 2D character array mappings where mappings[i] = [old_i, new_i] indicates that you may perform the following operation any number of times:

Replace a character old_i of sub with new_i.

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 <= 5000
0 <= mappings.length <= 1000
mappings[i].length == 2
old_i != new_i
s and sub consist of uppercase and lowercase English letters and digits.
old_i and new_i 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.

Python3JavaC++Go

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

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;
    }
}

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;
    }
};

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
}

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 $O(C^2)$.