LeetCode #1807 — MEDIUM

Evaluate the Bracket Pairs of a String

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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] = [key_i, value_i] indicates that key key_i has a value of value_i.

You are tasked to evaluate all of the bracket pairs. When you evaluate a bracket pair that contains some key key_i, you will:

Replace key_i and the bracket pair with the key's corresponding value_i.
If you do not know the value of the key, you will replace key_i 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 <= key_i.length, value_i.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.
key_i and value_i consist of lowercase English letters.
Each key_i in knowledge is unique.

Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

"(name)is(age)yearsold"
[["name","bob"],["age","two"]]

Example 2

"hi(name)"
[["a","b"]]

Example 3

"(a)(a)(a)aaa"
[["a","yes"]]

Core Insight

What unlocks the optimal approach

Process pairs from right to left to handle repeats
Keep track of the current enclosed string using another string

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1807: Evaluate the Bracket Pairs of a String
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();
    }
}

// Accepted solution for LeetCode #1807: Evaluate the Bracket Pairs of a String
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()
}

# Accepted solution for LeetCode #1807: Evaluate the Bracket Pairs of a String
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)

// Accepted solution for LeetCode #1807: Evaluate the Bracket Pairs of a String
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
    }
}

// Accepted solution for LeetCode #1807: Evaluate the Bracket Pairs of a 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('');
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n + m)

Space

O(L)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.