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Slowest Key

A newly designed keypad was tested, where a tester pressed a sequence of n keys, one at a time.

You are given a string keysPressed of length n, where keysPressed[i] was the ith key pressed in the testing sequence, and a sorted list releaseTimes, where releaseTimes[i] was the time the ith key was released. Both arrays are 0-indexed. The 0th key was pressed at the time 0, and every subsequent key was pressed at the exact time the previous key was released.

The tester wants to know the key of the keypress that had the longest duration. The ith keypress had a duration of releaseTimes[i] - releaseTimes[i - 1], and the 0th keypress had a duration of releaseTimes[0].

Note that the same key could have been pressed multiple times during the test, and these multiple presses of the same key may not have had the same duration.

Return the key of the keypress that had the longest duration. If there are multiple such keypresses, return the lexicographically largest key of the keypresses.

 

Example 1:

Input: releaseTimes = [9,29,49,50], keysPressed = "cbcd"
Output: "c"
Explanation: The keypresses were as follows:
Keypress for 'c' had a duration of 9 (pressed at time 0 and released at time 9).
Keypress for 'b' had a duration of 29 - 9 = 20 (pressed at time 9 right after the release of the previous character and released at time 29).
Keypress for 'c' had a duration of 49 - 29 = 20 (pressed at time 29 right after the release of the previous character and released at time 49).
Keypress for 'd' had a duration of 50 - 49 = 1 (pressed at time 49 right after the release of the previous character and released at time 50).
The longest of these was the keypress for 'b' and the second keypress for 'c', both with duration 20.
'c' is lexicographically larger than 'b', so the answer is 'c'.

Example 2:

Input: releaseTimes = [12,23,36,46,62], keysPressed = "spuda"
Output: "a"
Explanation: The keypresses were as follows:
Keypress for 's' had a duration of 12.
Keypress for 'p' had a duration of 23 - 12 = 11.
Keypress for 'u' had a duration of 36 - 23 = 13.
Keypress for 'd' had a duration of 46 - 36 = 10.
Keypress for 'a' had a duration of 62 - 46 = 16.
The longest of these was the keypress for 'a' with duration 16.

 

Constraints:

  • releaseTimes.length == n
  • keysPressed.length == n
  • 2 <= n <= 1000
  • 1 <= releaseTimes[i] <= 109
  • releaseTimes[i] < releaseTimes[i+1]
  • keysPressed contains only lowercase English letters.

Solution Explanation

The problem asks to find the key with the longest duration pressed. Duration is calculated as the difference between consecutive release times, or simply the release time for the first key. If multiple keys have the same longest duration, the lexicographically largest key should be returned.

Approach

The solution iterates through the releaseTimes and keysPressed arrays. For each keypress, it calculates the duration. It keeps track of the current longest duration (mx) and the corresponding key (ans). If a keypress has a longer duration than mx, or if it has the same duration but is lexicographically larger than the current ans, it updates mx and ans.

Time Complexity Analysis

The solution involves a single loop iterating through the releaseTimes and keysPressed arrays, each of length n. Therefore, the time complexity is O(n), which is linear with respect to the input size.

Space Complexity Analysis

The solution uses a constant amount of extra space to store variables like ans and mx. Therefore, the space complexity is O(1), which is constant.

Code Explanation (Python3 as an example, but the logic is identical across all languages)

class Solution:
    def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:
        ans = keysPressed[0]  # Initialize the answer with the first key
        mx = releaseTimes[0]  # Initialize the maximum duration with the release time of the first key
 
        for i in range(1, len(keysPressed)):  # Iterate from the second keypress
            d = releaseTimes[i] - releaseTimes[i - 1]  # Calculate the duration
            if d > mx or (d == mx and ord(keysPressed[i]) > ord(ans)):
                # Check if the duration is longer, or if it's equal and the key is lexicographically larger
                mx = d  # Update the maximum duration
                ans = keysPressed[i]  # Update the answer with the current key
 
        return ans  # Return the key with the longest duration

The ord() function is used to compare the keys lexicographically by comparing their ASCII values. The rest of the code directly implements the logic described in the approach section. The other language implementations follow the same logic with syntactic differences only.