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A laptop keyboard is connected to a computer. The computer records which keys are pressed, at what times and for how long the keys are held down. This data is stored in two lists:

• releaseTimes: contains the times at which a key is released.
• keysPressed: contains the corresponding keys that are pressed.

We are interested in finding the key with the longest duration that was pressed. If there is a tie, we need to return the lexicographically larger key.

Example

Input: releaseTimes = [9,29,49,50], keysPressed = "cbcd"
Output: "c"
Explanation: The keypress durations are as follows:
- 'c' -> 9 (from 0 to 9)
- 'b' -> 20 (from 9 to 29)
- 'c' -> 20 (from 29 to 49)
- 'd' -> 1 (from 49 to 50)
The longest duration is '20' and the lexicographically largest key is 'c'.

Clarifying Questions

1. What if releaseTimes and keysPressed are not of the same length?
   • Assume releaseTimes and keysPressed are always of the same length.
2. What is the maximum length of keysPressed?
   • The problem constraints usually handle a length up to 10^4, but it should be confirmed in the problem statement.

Strategy

1. Parse through the releaseTimes and keysPressed simultaneously.
2. Calculate the duration of each key press by subtracting the previous release time from the current release time.
3. Track the maximum duration found and the corresponding key.
4. If a key press duration ties with the maximum found duration, select the lexicographically larger key.

Code

def slowestKey(releaseTimes, keysPressed):
    max_duration = releaseTimes[0]
    slowest_key = keysPressed[0]
    
    for i in range(1, len(releaseTimes)):
        current_duration = releaseTimes[i] - releaseTimes[i-1]
        if current_duration > max_duration or (current_duration == max_duration and keysPressed[i] > slowest_key):
            max_duration = current_duration
            slowest_key = keysPressed[i]
    
    return slowest_key

# Example usage:
releaseTimes = [9, 29, 49, 50]
keysPressed = "cbcd"
print(slowestKey(releaseTimes, keysPressed))  # Output: "c"

Time Complexity

The strategy involves a single pass through the releaseTimes and keysPressed arrays. Given n as the length of the input lists:

• Time Complexity: O(n)
• Space Complexity: O(1) assuming the input list fits in memory since we’re only tracking a few extra variables for the maximum duration and corresponding key.

This solution should efficiently handle up to the maximum constraints typically given in such problems.

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