How do you solve Merge Close Characters on LeetCode?

Merge Close Characters (LeetCode #3853) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Merging depends on proximity and equality of characters.

What is the time complexity of Merge Close Characters?

The optimal solution for Merge Close Characters runs in O(n) time and uses O(n) space. Each character is processed once, and the stack operations are efficient.

What is the brute force solution for Merge Close Characters?

The brute force approach for Merge Close Characters is Brute Force, with O(n²) time complexity and O(1) space complexity. We check every pair of characters in the string to see if they can merge. This involves repeatedly scanning the string until no more merges are possible. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Merge Close Characters?

Merge Close Characters uses the following concepts: Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Merge Close Characters?

Clarify the merging rules before coding. Discuss edge cases like empty strings or no merges. Think about space and time complexity early.

What are common mistakes when solving Merge Close Characters?

Not considering the distance k correctly. Failing to update the string after each merge.

#3853

Merge Close Characters

Medium

Hash TableStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use a single pass with a stack to keep track of characters and their indices. This allows efficient merging by checking the top of the stack.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty stack.
2Step 2: For each character, check if it can merge with the top of the stack (within k distance).
3Step 3: If it can merge, pop the stack; otherwise, push the character onto the stack.

solution.py8 lines

1def mergeCloseCharacters(s, k):
2    stack = []
3    for i, char in enumerate(s):
4        if stack and stack[-1][0] == char and i - stack[-1][1] <= k:
5            stack.pop()
6        else:
7            stack.append((char, i))
8    return ''.join(c[0] for c in stack)

ℹ

Complexity note: Each character is processed once, and the stack operations are efficient.

1Merging depends on proximity and equality of characters.
2Using a stack allows efficient tracking of characters and their indices.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.