Leetcode Problem 1629. Slowest Key

1629. Slowest Key

Leetcode Solutions

Initialize slowestKey to the first key in keysPressed and longestPress to releaseTimes[0].
Iterate over releaseTimes starting from the second element.
Calculate currentDuration as the difference between the current and previous releaseTimes.
If currentDuration is greater than longestPress, update slowestKey and longestPress.
If currentDuration is equal to longestPress and the current key is lexicographically larger than slowestKey, update slowestKey.
Return slowestKey after the iteration is complete.

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