Class Solution
- java.lang.Object
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- g2301_2400.s2301_match_substring_after_replacement.Solution
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public class Solution extends Object
2301 - Match Substring After Replacement.Hard
You are given two strings
s
andsub
. You are also given a 2D character arraymappings
wheremappings[i] = [oldi, newi]
indicates that you may replace any number ofoldi
characters ofsub
withnewi
. Each character insub
cannot be replaced more than once.Return
true
if it is possible to makesub
a substring ofs
by replacing zero or more characters according tomappings
. Otherwise, returnfalse
.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
andsub
consist of uppercase and lowercase English letters and digits.oldi
andnewi
are either uppercase or lowercase English letters or digits.
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Constructor Summary
Constructors Constructor Description Solution()
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Method Summary
All Methods Instance Methods Concrete Methods Modifier and Type Method Description boolean
matchReplacement(String s, String sub, char[][] mappings)
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