Faulty Keyboard - Practice Coding Problems

Faulty Keyboard - Problem

String Simulation Easy

Imagine you're working on an old laptop with a quirky keyboard bug - every time you press the letter 'i', instead of just typing it, the keyboard reverses your entire string that you've written so far!

You're given a string s representing the sequence of characters you want to type. Your task is to simulate this faulty keyboard behavior and determine what the final string will look like on your screen.

The Rules:

When you type any character other than 'i' → it gets appended normally
When you type the character 'i' → the entire current string gets reversed

Goal: Return the final string that appears on your laptop screen after typing all characters in s.

Example: If you type "string", you get: s → st → str → rts (reverse after 'i') → rtsn → rtsng → final result: "rtsng"

Input & Output

example_1.py — Basic Case

$ Input: s = "string"

› Output: "rtsng"

💡 Note: Step by step: '' → 's' → 'st' → 'str' → (reverse on 'i') 'rts' → 'rtsn' → 'rtsng'

example_2.py — Multiple Reversals

$ Input: s = "poiinter"

› Output: "ponter"

💡 Note: '' → 'p' → 'po' → (reverse) 'op' → (reverse) 'po' → 'pon' → 'pont' → 'ponte' → 'ponter'

example_3.py — No 'i' Characters

$ Input: s = "hello"

› Output: "hello"

💡 Note: Since there are no 'i' characters, the string is built normally without any reversals

Constraints

1 ≤ s.length ≤ 100
s consists of only lowercase English letters
The character 'i' will trigger the reversal operation

Visualization

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Understanding the Visualization

Initialize

Start with empty screen

Type Normal Chars

Characters other than 'i' get appended normally

Hit 'i' Key

The magic happens! Entire string gets reversed

Continue Typing

Keep building on the reversed string

Final Result

Get the final string after all characters processed

Key Takeaway

🎯 Key Insight: Each 'i' character acts as a reverse operation on the current string, making this a straightforward simulation problem

Asked in

f Meta 15 G Google 12 a Amazon 8 ⊞ Microsoft 6

The optimal approach uses string simulation with character-by-character processing. We iterate through the input string, appending normal characters and reversing the entire result whenever we encounter the character 'i'. This directly mirrors the faulty keyboard behavior with O(n²) time complexity in worst case due to string reversals.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force String Reversal	O(n²)	O(n)	Build result string character by character, reversing the entire string each time we encounter 'i'

Brute Force String Reversal — Algorithm Steps

Initialize an empty result string
Iterate through each character in the input string
If the character is not 'i', append it to the result
If the character is 'i', reverse the entire result string
Return the final result string

Visualization

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Step-by-Step Walkthrough

Process 's'

Add 's' to empty string: result = 's'

Process 't'

Add 't' to result: result = 'st'

Process 'r'

Add 'r' to result: result = 'str'

Process 'i'

Reverse the entire string: result = 'rts'

Process 'n'

Add 'n' to result: result = 'rtsn'

Process 'g'

Add 'g' to result: result = 'rtsng'

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

void reverseString(char* str) {
    int len = strlen(str);
    for (int i = 0; i < len / 2; i++) {
        char temp = str[i];
        str[i] = str[len - 1 - i];
        str[len - 1 - i] = temp;
    }
}

char* finalString(char* s) {
    int len = strlen(s);
    char* result = (char*)malloc((len + 1) * sizeof(char));
    result[0] = '\0';
    
    for (int i = 0; i < len; i++) {
        if (s[i] == 'i') {
            reverseString(result);
        } else {
            int resLen = strlen(result);
            result[resLen] = s[i];
            result[resLen + 1] = '\0';
        }
    }
    
    return result;
}

int main() {
    char input[1001];
    fgets(input, sizeof(input), stdin);
    input[strcspn(input, "\n")] = '\0';
    
    // Remove quotes
    char* s = input + 1;
    s[strlen(s) - 1] = '\0';
    
    char* result = finalString(s);
    printf("\"%s\"", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

In worst case with many 'i' characters, we reverse the string multiple times, each reversal taking O(n) time

⚠ Quadratic Growth

Space Complexity

O(n)

We store the result string which can be at most n characters long

⚡ Linearithmic Space

23.4K Views

Medium Frequency

~8 min Avg. Time

890 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force String Reversal — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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