How do you solve Faulty Keyboard on LeetCode?

Faulty Keyboard (LeetCode #2810) 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: Reversing a string is costly; find ways to minimize reversals.

What is the brute force solution for Faulty Keyboard?

The brute force approach for Faulty Keyboard is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach processes each character in the string one by one, updating the result string. When encountering the character 'i', it reverses the current string. This is straightforward but inefficient due to repeated reversals. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Faulty Keyboard?

Faulty Keyboard uses the following concepts: String, Simulation. The recommended patterns to study are: Stack, Two Pointers.

What are the interview tips for Faulty Keyboard?

Think about how to minimize costly operations like reversals. Explain your thought process clearly while coding. Consider edge cases, like empty strings or strings with only 'i's.

What are common mistakes when solving Faulty Keyboard?

Not handling multiple consecutive 'i's properly. Forgetting to reverse the final stack before returning.

#2810

Faulty Keyboard

Easy

StringSimulationStackTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Instead of reversing the string every time we encounter 'i', we can use a stack to keep track of the characters. When we hit 'i', we can simply reverse the stack instead of the entire string, which is more efficient.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty stack.
2Step 2: Iterate through each character in the input string.
3Step 3: If the character is 'i', reverse the stack; otherwise, push the character onto the stack.
4Step 4: After processing all characters, join the stack to form the final string.

solution.py8 lines

1def faulty_keyboard(s):
2    stack = []
3    for char in s:
4        if char == 'i':
5            stack.reverse()
6        else:
7            stack.append(char)
8    return ''.join(stack)

ℹ

Complexity note: The time complexity is O(n) because we process each character once. The space complexity is O(n) due to the stack used to store the characters.

1Reversing a string is costly; find ways to minimize reversals.
2Using a stack allows for efficient character management.

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