Slowest Key

class Solution {
public:
    char findSlowestKey(vector<int>& releaseTimes, string keysPressed) {
        int numberOfKeys = releaseTimes.size();
        int maxDuration = releaseTimes[0];
        char resultKey = keysPressed[0];
        for (int i = 1; i < numberOfKeys; i++) {
            int duration = releaseTimes[i] - releaseTimes[i - 1];
            if (duration > maxDuration ||
                (duration == maxDuration && keysPressed[i] > resultKey)) {
                maxDuration = duration;
                resultKey = keysPressed[i];
            }
        }
        return resultKey;
    }
};

This solution, implemented in C++, is designed to determine the slowest key from a series of key presses. It involves calculating the longest duration a key remains pressed based on timing data provided for each key release.

Follow these steps in the provided code to understand how to find the slowest key:

1. Initialize numberOfKeys with the size of releaseTimes to get the total number of key presses.
2. Start by assigning maxDuration to the duration of the first key press (releaseTimes[0]), and resultKey to the first key in keysPressed.
3. Iterate through the releaseTimes starting from the second key press (index 1).
4. Calculate the duration for the current key by subtracting the previous release time from the current one.
5. Compare the duration of the current key with maxDuration.
   • If the duration is greater, update maxDuration and set resultKey to the current key.
   • If the duration is equal to the maxDuration, further compare the keys. Update resultKey if the current key is lexicographically greater.
6. Once the loop completes, resultKey will store the character of the slowest key.

This approach ensures that you efficiently find the key with the longest press time, considering all edge cases, such as when two keys have the same duration.

class Solution {
    
    public char findSlowestKey(int[] pressTimes, String keySequence) {
        int count = pressTimes.length;
        int maxDuration = pressTimes[0];
        char resultKey = keySequence.charAt(0);
        for (int index = 1; index < count; index++) {
            int duration = pressTimes[index] - pressTimes[index - 1];
            // Determine if the current key has the longest press
            if (duration > maxDuration ||
                (duration == maxDuration && keySequence.charAt(index) > resultKey)) {
                maxDuration = duration;
                resultKey = keySequence.charAt(index);
            }
        }
        return resultKey;
    }
}

The provided Java solution efficiently determines the slowest key from a sequence of key presses. The method findSlowestKey accepts two parameters: pressTimes, an array of integers where each integer represents the time at which a key is pressed, and keySequence, a string where each character represents a specific key pressed at the corresponding time in the pressTimes array.

• Initialize count to the length of the pressTimes array.
• Set maxDuration to the duration of the first key pressed, as indicated by the first element in the pressTimes.
• Set resultKey to the first character of the keySequence.
• Loop through each subsequent key press starting from the second. Calculate the duration of each key press by subtracting the press time of the previous key from the current key's press time.
• Compare the calculated duration with the recorded maxDuration. If the current duration is greater or if it is the same but the key itself has a higher lexicographical order than the current slowest key, update maxDuration and resultKey.
• Finally, return the key (resultKey) with the longest press time or the lexicographically greater key in case of tie durations, which is considered the slowest.

This solution effectively utilizes a for-loop and conditional statements to compare each key's press duration and ensures the correct slowest key is identified with a time complexity of O(n), where n is the number of keys pressed.