public class Solution extends Object
2301 - Match Substring After Replacement.
Hard
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 replace any number of oldi
characters 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 <= 5000
0 <= mappings.length <= 1000
mappings[i].length == 2
oldi != newi
s
and sub
consist of uppercase and lowercase English letters and digits.oldi
and newi
are either uppercase or lowercase English letters or digits.Constructor and Description |
---|
Solution() |
Modifier and Type | Method and Description |
---|---|
boolean |
matchReplacement(String s,
String sub,
char[][] mappings) |
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