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.@Suppress("NAME_SHADOWING")
class Solution {
private lateinit var c1: CharArray
private lateinit var c2: CharArray
private lateinit var al: Array<MutableSet<Char>?>
fun matchReplacement(s: String, sub: String, mappings: Array<CharArray>): Boolean {
c1 = s.toCharArray()
c2 = sub.toCharArray()
al = arrayOfNulls(75)
for (i in 0..74) {
val temp: MutableSet<Char> = HashSet()
al[i] = temp
}
for (mapping in mappings) {
al[mapping[0].code - '0'.code]!!.add(mapping[1])
}
return ans(c1.size, c2.size) == 1
}
private fun ans(m: Int, n: Int): Int {
var m = m
var n = n
if (m == 0) {
return 0
}
if (ans(m - 1, n) == 1) {
return 1
}
if (m >= n && (c1[m - 1] == c2[n - 1] || al[c2[n - 1].code - '0'.code]!!.contains(c1[m - 1]))) {
while (n >= 1 && (c1[m - 1] == c2[n - 1] || al[c2[n - 1].code - '0'.code]!!.contains(c1[m - 1]))) {
n--
m--
}
if (n == 0) {
return 1
}
}
return 0
}
}