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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.
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.
1using System;
2
3public class SlowestKeySolution {
4 public char SlowestKey(int[] releaseTimes, string keysPressed) {
5 int maxDuration = releaseTimes[0];
6 char result = keysPressed[0];
7 for (int i = 1; i < releaseTimes.Length; i++) {
8 int duration = releaseTimes[i] - releaseTimes[i - 1];
9 if (duration > maxDuration || (duration == maxDuration && keysPressed[i] > result)) {
10 maxDuration = duration;
11 result = keysPressed[i];
12 }
13 }
14 return result;
15 }
16
17 public static void Main() {
18 SlowestKeySolution solution = new SlowestKeySolution();
19 Console.WriteLine(solution.SlowestKey(new int[] { 12, 23, 36, 46, 62 }, "spuda"));
20 }
21}This C# implementation is methodically structured, mirroring the logic laid out in previous languages with clear and concise iteration over the input arrays.
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.
Time Complexity: O(n)
Space Complexity: O(n) due to additional array usage.
1function
In this JavaScript implementation, a temporary durations array helps facilitate the separation of computation and decision processes.