Description
Brace Expansion II
HardUnder 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:
- Single letters represent a singleton set containing that word.
R("a") = {"a"}R("w") = {"w"}
- When we take a comma-delimited list of two or more expressions, we take the union of possibilities.
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)
- When we concatenate two expressions, we take the set of possible concatenations between two words where the first word comes from the first expression and the second word comes from the second expression.
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:
- For every lowercase letter
x, we haveR(x) = {x}. - For expressions
e1, e2, ... , ekwithk >= 2, we haveR({e1, e2, ...}) = R(e1) ∪ R(e2) ∪ ... - For expressions
e1ande2, we haveR(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 <= 60expression[i]consists of'{','}',','or lowercase English letters.- The given
expressionrepresents a set of words based on the grammar given in the description.
Solutions
This seemed like a fun exercise that I might have tackled on a weekend-day.
Language: javascript()DONE
CPU Performance95.24%
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Memory Performance100.00%
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/** By Leetcode */
var braceExpansionII = function (expression) {
const op = []; // Operator stack
const stk = []; // Set stack
// Pop the operator at the top of the stack and perform the calculation
const ope = () => {
const l = stk.length - 2,
r = stk.length - 1;
if (op[op.length - 1] === "+") {
// Union operation
for (const item of stk[r]) {
stk[l].add(item);
}
} else {
// Cartesian product operation
const tmp = new Set();
for (const left of stk[l]) {
for (const right of stk[r]) {
tmp.add(left + right);
}
}
stk[l] = tmp;
}
op.pop();
stk.pop();
};
for (let i = 0; i < expression.length; i++) {
const ch = expression[i];
if (ch === ",") {
// Keep popping operators from the top of the stack until the stack is empty or its top is not a multiplication sign
while (op.length > 0 && op[op.length - 1] === "*") {
ope();
}
op.push("+");
} else if (ch === "{") {
// First determine whether a multiplication sign needs to be added, then push { onto the operator stack
if (
i > 0 &&
(expression[i - 1] === "}" || /[a-z]/.test(expression[i - 1]))
) {
op.push("*");
}
op.push("{");
} else if (ch === "}") {
// Keep popping operators from the top of the stack until its top is {
while (op.length > 0 && op[op.length - 1] !== "{") {
ope();
}
op.pop();
} else {
// First determine whether a multiplication sign needs to be added, then push the newly constructed set onto the set stack
if (
i > 0 &&
(expression[i - 1] === "}" || /[a-z]/.test(expression[i - 1]))
) {
op.push("*");
}
stk.push(new Set([ch]));
}
}
while (op.length > 0) {
ope();
}
return Array.from(stk[stk.length - 1]).sort();
};