LeetCode #2810 — EASY

Faulty Keyboard

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

Your laptop keyboard is faulty, and whenever you type a character 'i' on it, it reverses the string that you have written. Typing other characters works as expected.

You are given a 0-indexed string s, and you type each character of s using your faulty keyboard.

Return the final string that will be present on your laptop screen.

Example 1:

Input: s = "string"
Output: "rtsng"
Explanation: 
After typing first character, the text on the screen is "s".
After the second character, the text is "st". 
After the third character, the text is "str".
Since the fourth character is an 'i', the text gets reversed and becomes "rts".
After the fifth character, the text is "rtsn". 
After the sixth character, the text is "rtsng". 
Therefore, we return "rtsng".

Example 2:

Input: s = "poiinter"
Output: "ponter"
Explanation: 
After the first character, the text on the screen is "p".
After the second character, the text is "po". 
Since the third character you type is an 'i', the text gets reversed and becomes "op". 
Since the fourth character you type is an 'i', the text gets reversed and becomes "po".
After the fifth character, the text is "pon".
After the sixth character, the text is "pont". 
After the seventh character, the text is "ponte". 
After the eighth character, the text is "ponter". 
Therefore, we return "ponter".

Constraints:

1 <= s.length <= 100
s consists of lowercase English letters.
s[0] != 'i'

Step 01

Brute Force Baseline

Problem summary: Your laptop keyboard is faulty, and whenever you type a character 'i' on it, it reverses the string that you have written. Typing other characters works as expected. You are given a 0-indexed string s, and you type each character of s using your faulty keyboard. Return the final string that will be present on your laptop screen.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"string"

Example 2

"poiinter"

Core Insight

What unlocks the optimal approach

Try to build a new string by traversing the given string and reversing whenever you get the character ‘i’.

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 #2810: Faulty Keyboard
class Solution {
    public String finalString(String s) {
        StringBuilder t = new StringBuilder();
        for (char c : s.toCharArray()) {
            if (c == 'i') {
                t.reverse();
            } else {
                t.append(c);
            }
        }
        return t.toString();
    }
}

// Accepted solution for LeetCode #2810: Faulty Keyboard
func finalString(s string) string {
	t := []rune{}
	for _, c := range s {
		if c == 'i' {
			for i, j := 0, len(t)-1; i < j; i, j = i+1, j-1 {
				t[i], t[j] = t[j], t[i]
			}
		} else {
			t = append(t, c)
		}
	}
	return string(t)
}

# Accepted solution for LeetCode #2810: Faulty Keyboard
class Solution:
    def finalString(self, s: str) -> str:
        t = []
        for c in s:
            if c == "i":
                t = t[::-1]
            else:
                t.append(c)
        return "".join(t)

// Accepted solution for LeetCode #2810: Faulty Keyboard
impl Solution {
    pub fn final_string(s: String) -> String {
        let mut t = Vec::new();
        for c in s.chars() {
            if c == 'i' {
                t.reverse();
            } else {
                t.push(c);
            }
        }
        t.into_iter().collect()
    }
}

// Accepted solution for LeetCode #2810: Faulty Keyboard
function finalString(s: string): string {
    const t: string[] = [];
    for (const c of s) {
        if (c === 'i') {
            t.reverse();
        } else {
            t.push(c);
        }
    }
    return t.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^2)

Space

O(n)

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.