Leetcode 2810: Faulty Keyboard

Input: A string s consisting of lowercase English letters, where the first character is not 'i'.

Example: Input: s = "hello"

Constraints:

• 1 <= s.length <= 100

• s consists of lowercase English letters

• s[0] != 'i'

Output: Return the final string that will be displayed on the screen after typing all characters in the string s.

Example: Output: "hll"

Constraints:

Goal: Simulate the process of typing each character of the string on a faulty keyboard, where typing 'i' reverses the string.

Steps:

• 1. Initialize an empty string to simulate the screen content.

• 2. For each character in the string s, add it to the string unless it is 'i'.

• 3. If the character is 'i', reverse the string typed so far.

Goal: Ensure that the solution can handle strings up to the maximum length efficiently.

Steps:

• 1 <= s.length <= 100

• s consists of lowercase English letters

• s[0] != 'i'

Assumptions:

• The string length will not exceed 100 characters.

• The input string will always begin with a character other than 'i'.

• Input: Input: s = "hello"

• Explanation: Typing the characters one by one and applying the reverse operation on encountering 'i' leads to the final string 'hll'.

• Input: Input: s = "worldi"

• Explanation: After typing 'world' and then encountering 'i', the string is reversed to 'dworl'.

Approach: We will simulate the process of typing the string and apply the reverse operation when encountering 'i'.

Observations:

• We only need to reverse the string when 'i' is encountered, which makes the problem manageable.

• We need to ensure that the string is updated after every character input, and we reverse only when needed.

Steps:

• 1. Start with an empty string representing the screen content.

• 2. For each character in the input string, either add it normally to the screen or reverse the string if 'i' is encountered.

Empty Inputs:

• There will always be at least one character in the string.

Large Inputs:

• Ensure the solution works efficiently for strings of length up to 100.

Special Values:

• If the string contains no 'i', the output will simply be the same as the input string.

Constraints:

• The solution must handle the string reversal operation efficiently.

string finalString(string s) {
    string res = "";
    for(auto c : s){
        if(c == 'i'){
            if(res.size()){
                reverse(res.begin(), res.end());
            }
            
        }else
         res+=c;
    }
    return res;
}

1 : Function Definition

string finalString(string s) {

The function 'finalString' takes a string 's' as input and returns a string as output after processing.

2 : Variable Initialization

    string res = "";

Initialize an empty string 'res' to store the modified characters of the input string.

3 : Looping

    for(auto c : s){

Iterate through each character 'c' in the input string 's'.

4 : Condition Check

        if(c == 'i'){

Check if the current character 'c' is 'i'. If true, the string 'res' will be reversed.

5 : Condition Check

            if(res.size()){

Check if the string 'res' is non-empty before attempting to reverse it.

6 : String Reversal

                reverse(res.begin(), res.end());

Reverse the contents of the string 'res' using the 'reverse' function from the C++ standard library.

7 : Condition Check

        }else

If the character is not 'i', continue processing the character by adding it to 'res'.

8 : String Concatenation

         res+=c;

Append the current character 'c' to the string 'res'.

9 : Return Statement

    return res;

Return the modified string 'res' after all characters have been processed.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: In the worst case, we reverse the string on each 'i' encounter, leading to a linear time complexity.

Best Case: O(n)

Worst Case: O(n)

Description: We store the result string which can grow up to the size of the input string.

Leetcode 2810: Faulty Keyboard

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