Hard
Under the grammar given below, strings can represent a set of lowercase words. Let R(expr)
denote the set of words the expression represents.
The grammar can best be understood through simple examples:
R("a") = {"a"}
R("w") = {"w"}
R("{a,b,c}") = {"a","b","c"}
R("{ {a,b},{b,c}}") = {"a","b","c"}
(notice the final set only contains each word at most once)R("{a,b}{c,d}") = {"ac","ad","bc","bd"}
R("a{b,c}{d,e}f{g,h}") = {"abdfg", "abdfh", "abefg", "abefh", "acdfg", "acdfh", "acefg", "acefh"}
Formally, the three rules for our grammar:
x
, we have R(x) = {x}
.e1, e2, ... , ek
with k >= 2
, we have R({e1, e2, ...}) = R(e1) ∪ R(e2) ∪ ...
e1
and e2
, we have R(e1 + e2) = {a + b for (a, b) in R(e1) × R(e2)}
, where +
denotes concatenation, and ×
denotes the cartesian product.Given an expression representing a set of words under the given grammar, return the sorted list of words that the expression represents.
Example 1:
Input: expression = “{a,b}{c,{d,e}}”
Output: [“ac”,”ad”,”ae”,”bc”,”bd”,”be”]
Example 2:
Input: expression = “{ {a,z},a{b,c},{ab,z}}”
Output: [“a”,”ab”,”ac”,”z”]
Explanation: Each distinct word is written only once in the final answer.
Constraints:
1 <= expression.length <= 60
expression[i]
consists of '{'
, '}'
, ','
or lowercase English letters.expression
represents a set of words based on the grammar given in the description.class Solution {
fun braceExpansionII(expression: String): List<String> {
val res = flatten(expression)
val sorted: MutableList<String> = ArrayList(res)
sorted.sort()
return sorted
}
private fun flatten(expression: String): Set<String> {
val res: MutableSet<String> = HashSet()
// A temp set to store cartesian product results.
var curSet: MutableSet<String> = HashSet()
var idx = 0
while (idx < expression.length) {
if (expression[idx] == '{') {
// end will be the index of matching "}"
val end = findClosingBrace(expression, idx)
val set = flatten(expression.substring(idx + 1, end))
curSet = concatenateSet(curSet, set)
idx = end + 1
} else if (Character.isLowerCase(expression[idx])) {
// Create set with single element
val set: Set<String> = HashSet(
listOf(
expression[idx].toString(),
),
)
curSet = concatenateSet(curSet, set)
idx++
} else if (expression[idx] == ',') {
res.addAll(curSet)
curSet.clear()
idx++
}
}
// Don't forget!
res.addAll(curSet)
return res
}
private fun concatenateSet(set1: Set<String>, set2: Set<String>): MutableSet<String> {
if (set1.isEmpty() || set2.isEmpty()) {
return if (set2.isNotEmpty()) HashSet(set2) else HashSet(set1)
}
val res: MutableSet<String> = HashSet()
for (s1 in set1) {
for (s2 in set2) {
res.add(s1 + s2)
}
}
return res
}
private fun findClosingBrace(expression: String, start: Int): Int {
var count = 0
var idx = start
while (idx < expression.length) {
if (expression[idx] == '{') {
count++
} else if (expression[idx] == '}') {
count--
}
if (count == 0) {
break
}
idx++
}
return idx
}
}