Faulty Keyboard

Easy

Samsung

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Topics:

StringsTwo Pointers

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'

Faulty Keyboard

Easy

Samsung

0 views

Topics:

StringsTwo Pointers

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'

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible characters that can appear in the input string s?
Can the input string s be empty or null?
If multiple corrections are possible, is any corrected string acceptable, or is there a specific correction that's preferred?
Is the input case-sensitive? Should 'I' and 'i' be treated the same?
Are there any other faulty key behaviors besides the double 'i' press that I should be aware of?

Brute Force Solution

Approach

The brute force approach to the Faulty Keyboard problem involves checking every possible interpretation of the input string. We'll assume each character could have been typed correctly or incorrectly (reversed) and try every combination. It's like trying all the different possibilities to see which one matches what we expect.

Here's how the algorithm would work step-by-step:

Start by assuming the first character was typed correctly, and build a string based on that.
Then, assume the first character was typed incorrectly (reversed), and build another string.
For each of these strings, consider the second character. Again, assume it's typed correctly and build a new string. Then assume it's typed incorrectly and build yet another string.
Keep doing this for every character in the input. You'll end up with a huge number of possibilities.
Compare each of these possibilities against the expected output. If a possibility matches the expected output, then it's a valid solution.
If there are multiple valid solutions, we can keep track of them or choose the first one we find, depending on what the problem asks.

Code Implementation

def faulty_keyboard_brute_force(faulty_word, target_word):
    # Generates all possible combinations of reversed segments
    def generate_combinations(current_index, current_string):
        if current_index == len(faulty_word):
            combinations.append(current_string)
            return

        # Recursively calls to generate a forward segment of the word
        generate_combinations(current_index + 1, current_string + faulty_word[current_index])

        # Recursively calls to generate a reversed segment of the word
        reversed_segment = faulty_word[current_index]
        generate_combinations(current_index + 1, current_string + reversed_segment[::-1])

    combinations = []
    generate_combinations(0, "")

    # Check if any combination matches the target word.
    for combination in combinations:
        if combination == target_word:
            return True

    # If none of the combinations match, return False.
    return False

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach considers all possible interpretations of the input string, where each character can be either correctly typed or reversed. For an input string of length n, there are 2 possibilities for each character. Therefore, we explore 2*2*...*2 (n times) possible strings. This means the time complexity grows exponentially with the input size n, resulting in O(2^n).

Space Complexity

O(2^N) – The brute force approach generates all possible interpretations of the input string, where each character can be either correctly typed or reversed. This leads to a binary tree-like exploration with a depth of N (the length of the input string), where each level represents a character and the two branches represent the two possibilities. The algorithm essentially explores all 2^N paths, storing each intermediate string. Therefore, the space complexity is dominated by the storage of these intermediate strings, resulting in O(2^N) space complexity due to the exponential growth of possibilities with the input size N. Each of these strings would have a maximum length of N.

Optimal Solution

Approach

The core idea is to simulate typing on the faulty keyboard and build the final string character by character. When we encounter a character, we decide whether it should be appended normally or flipped based on whether the keyboard is currently toggled on or off.

Here's how the algorithm would work step-by-step:

Start with an empty string which will hold the final output.
Keep track of whether the keyboard's 'invert' function is currently on or off (start with it off).
Go through the input string one character at a time.
If the character is 'i', flip the 'invert' function on or off.
If the keyboard's 'invert' function is on, 'flip' the character (e.g. turn 'a' into 'A' or vice versa) and add the flipped character to the final output string.
If the keyboard's 'invert' function is off, just add the original character to the final output string.
After processing all characters in the input, the final output string is the result.

Code Implementation

def faulty_keyboard(input_string):
    processed_string = ""
    reverse_flag = False

    for char in input_string:
        if char == 'i':
            # Toggle the flag when 'i' is encountered
            reverse_flag = not reverse_flag

        else:
            processed_string += char

    # Reverse the string if the reverse flag is true
    if reverse_flag:
        processed_string = processed_string[::-1]

    return processed_string

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input string of size n once. Inside the loop, the operations performed are constant time operations: checking the character, potentially flipping the invert state, potentially flipping the character case, and appending to the result string. Since the number of operations grows linearly with the input size n, the time complexity is O(n).

Space Complexity

O(N) – The algorithm constructs a new string to hold the final output, character by character. The size of this output string can grow up to the length of the input string if no 'i' characters are present, or potentially be larger if a significant portion of the characters are flipped. Therefore, the auxiliary space used is proportional to the length of the input string, denoted as N. Thus, the space complexity is O(N).

Edge Cases

Case	How to Handle
Input string is null or empty	Return an empty string to avoid null pointer exceptions or incorrect processing.
Input string contains only the character 'i'	Handle consecutive 'i's to ensure correct reduction of double 'i's to single 'i'.
Input string contains no 'i' characters	Return the original string, as no correction is needed.
Input string has consecutive 'i' characters at the beginning or end	The loop should handle boundary conditions without index-out-of-bounds errors.
Input string contains only repeated 'i' characters (e.g., 'iiiiiiii')	Reduce the string to half the number of 'i' characters, rounding up if necessary for odd lengths.
Input string contains mixed characters including single and double 'i's	Process the string correctly by iterating and reducing only doubled 'i's.
Maximum length string, check for performance	Iterative solutions are more efficient than recursive ones, particularly for large strings.
Input string has alternating single and double 'i's (e.g., 'i' 'ii' 'i' 'ii')	Ensure correct identification and reduction of only the double 'i' sequences while retaining single 'i's.