Class Solution
- java.lang.Object
-
- g1801_1900.s1807_evaluate_the_bracket_pairs_of_a_string.Solution
-
public class Solution extends Object
1807 - Evaluate the Bracket Pairs of a String.Medium
You are given a string
s
that contains some bracket pairs, with each pair containing a non-empty key.- For example, in the string
"(name)is(age)yearsold"
, there are two bracket pairs that contain the keys"name"
and"age"
.
You know the values of a wide range of keys. This is represented by a 2D string array
knowledge
where eachknowledge[i] = [keyi, valuei]
indicates that keykeyi
has a value ofvaluei
.You are tasked to evaluate all of the bracket pairs. When you evaluate a bracket pair that contains some key
keyi
, you will:- Replace
keyi
and the bracket pair with the key’s correspondingvaluei
. - If you do not know the value of the key, you will replace
keyi
and the bracket pair with a question mark"?"
(without the quotation marks).
Each key will appear at most once in your
knowledge
. There will not be any nested brackets ins
.Return the resulting string after evaluating all of the bracket pairs.
Example 1:
Input: s = “(name)is(age)yearsold”, knowledge = [[“name”,“bob”],[“age”,“two”]]
Output: “bobistwoyearsold”
Explanation:
The key “name” has a value of “bob”, so replace “(name)” with “bob”.
The key “age” has a value of “two”, so replace “(age)” with “two”.
Example 2:
Input: s = “hi(name)”, knowledge = [[“a”,“b”]]
Output: “hi?”
Explanation: As you do not know the value of the key “name”, replace “(name)” with “?”.
Example 3:
Input: s = “(a)(a)(a)aaa”, knowledge = [[“a”,“yes”]]
Output: “yesyesyesaaa”
Explanation: The same key can appear multiple times.
The key “a” has a value of “yes”, so replace all occurrences of “(a)” with “yes”.
Notice that the "a"s not in a bracket pair are not evaluated.
Constraints:
1 <= s.length <= 105
0 <= knowledge.length <= 105
knowledge[i].length == 2
1 <= keyi.length, valuei.length <= 10
s
consists of lowercase English letters and round brackets'('
and')'
.- Every open bracket
'('
ins
will have a corresponding close bracket')'
. - The key in each bracket pair of
s
will be non-empty. - There will not be any nested bracket pairs in
s
. keyi
andvaluei
consist of lowercase English letters.- Each
keyi
inknowledge
is unique.
- For example, in the string
-
-
Constructor Summary
Constructors Constructor Description Solution()
-