Question

1807. Evaluate the Bracket Pairs of a String

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Description

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 each knowledge[i] = [keyi, valuei] indicates that key keyi has a value of valuei.

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 corresponding valuei.
• 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 in s.

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 '(' in s 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 and valuei consist of lowercase English letters.
• Each keyi in knowledge is unique.

Solutions

Solution 1: Hash Table + Simulation

First, we use a hash table $d$ to record the key-value pairs in knowledge.

Then we traverse the string $s$. If the current character is an open parenthesis '(', we start traversing from the current position until we encounter a close parenthesis ')'. At this point, the string within the parentheses is the key. We look for the corresponding value of this key in the hash table $d$. If found, we replace the value within the parentheses with it, otherwise, we replace it with '?'.

The time complexity is $O(n + m)$, and the space complexity is $O(L)$. Here, $n$ and $m$ are the lengths of the string $s$ and the list knowledge respectively, and $L$ is the sum of the lengths of all strings in knowledge.

class Solution:
  def evaluate(self, s: str, knowledge: List[List[str]]) -> str:
    d = {a: b for a, b in knowledge}
    i, n = 0, len(s)
    ans = []
    while i < n:
      if s[i] == '(':
        j = s.find(')', i + 1)
        ans.append(d.get(s[i + 1 : j], '?'))
        i = j
      else:
        ans.append(s[i])
      i += 1
    return ''.join(ans)

class Solution {
  public String evaluate(String s, List<List<String>> knowledge) {
    Map<String, String> d = new HashMap<>(knowledge.size());
    for (var e : knowledge) {
      d.put(e.get(0), e.get(1));
    }
    StringBuilder ans = new StringBuilder();
    for (int i = 0; i < s.length(); ++i) {
      if (s.charAt(i) == '(') {
        int j = s.indexOf(')', i + 1);
        ans.append(d.getOrDefault(s.substring(i + 1, j), "?"));
        i = j;
      } else {
        ans.append(s.charAt(i));
      }
    }
    return ans.toString();
  }
}

class Solution {
public:
  string evaluate(string s, vector<vector<string>>& knowledge) {
    unordered_map<string, string> d;
    for (auto& e : knowledge) {
      d[e[0]] = e[1];
    }
    string ans;
    for (int i = 0; i < s.size(); ++i) {
      if (s[i] == '(') {
        int j = s.find(")", i + 1);
        auto t = s.substr(i + 1, j - i - 1);
        ans += d.count(t) ? d[t] : "?";
        i = j;
      } else {
        ans += s[i];
      }
    }
    return ans;
  }
};

func evaluate(s string, knowledge [][]string) string {
  d := map[string]string{}
  for _, v := range knowledge {
    d[v[0]] = v[1]
  }
  var ans strings.Builder
  for i := 0; i < len(s); i++ {
    if s[i] == '(' {
      j := i + 1
      for s[j] != ')' {
        j++
      }
      if v, ok := d[s[i+1:j]]; ok {
        ans.WriteString(v)
      } else {
        ans.WriteByte('?')
      }
      i = j
    } else {
      ans.WriteByte(s[i])
    }
  }
  return ans.String()
}

function evaluate(s: string, knowledge: string[][]): string {
  const n = s.length;
  const map = new Map();
  for (const [k, v] of knowledge) {
    map.set(k, v);
  }
  const ans = [];
  let i = 0;
  while (i < n) {
    if (s[i] === '(') {
      const j = s.indexOf(')', i + 1);
      ans.push(map.get(s.slice(i + 1, j)) ?? '?');
      i = j;
    } else {
      ans.push(s[i]);
    }
    i++;
  }
  return ans.join('');
}

use std::collections::HashMap;
impl Solution {
  pub fn evaluate(s: String, knowledge: Vec<Vec<String>>) -> String {
    let s = s.as_bytes();
    let n = s.len();
    let mut map = HashMap::new();
    for v in knowledge.iter() {
      map.insert(&v[0], &v[1]);
    }
    let mut ans = String::new();
    let mut i = 0;
    while i < n {
      if s[i] == b'(' {
        i += 1;
        let mut j = i;
        let mut key = String::new();
        while s[j] != b')' {
          key.push(s[j] as char);
          j += 1;
        }
        ans.push_str(map.get(&key).unwrap_or(&&'?'.to_string()));
        i = j;
      } else {
        ans.push(s[i] as char);
      }
      i += 1;
    }
    ans
  }
}