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 == nkeysPressed.length == n2 <= n <= 10001 <= releaseTimes[i] <= 109releaseTimes[i] < releaseTimes[i+1]keysPressed contains only lowercase English letters.This approach involves iterating through both releaseTimes and keysPressed to calculate the duration of each keypress. We'll track the maximum duration found so far and update our result each time we find a keypress with a longer duration. If two keypresses have the same duration, we choose the lexicographically larger key.
We initialize the maximum duration with the duration of the first keypress and iteratively check each subsequent duration. The key with the longest duration is tracked and returned. The code checks if the current duration is greater than the maximum duration found so far or if the key is lexicographically larger when durations are equal.
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Time Complexity: O(n), where n is the length of the releaseTimes array. We iterate through the list once.
Space Complexity: O(1), as we only use constant extra space for variables.
In this approach, we create a temporary array to store the duration of each key press. We then iterate over this array to find the maximum duration. This solution separates the concerns of calculation and comparison for more clarity.
This approach uses an array to store durations, which is iterated through twice. This can make the logic a bit clearer by separating computation and decision logic.
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Java
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JavaScript
Time Complexity: O(n)
Space Complexity: O(n) due to additional array usage.
| Approach | Complexity |
|---|---|
| Simple Iteration with Maximum Duration Tracking | Time Complexity: O(n), where n is the length of the releaseTimes array. We iterate through the list once. |
| Using Temporary Array for Duration Calculation | Time Complexity: O(n) |
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