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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.
1public class SlowestKey {
2 public char slowestKey(int[] releaseTimes, String keysPressed) {
3 int maxDuration = releaseTimes[0];
4 char result = keysPressed.charAt(0);
5 for (int i = 1; i < releaseTimes.length; i++) {
6 int duration = releaseTimes[i] - releaseTimes[i - 1];
7 if (duration > maxDuration || (duration == maxDuration && keysPressed.charAt(i) > result)) {
8 maxDuration = duration;
9 result = keysPressed.charAt(i);
10 }
11 }
12 return result;
13 }
14
15 public static void main(String[] args) {
16 SlowestKey solution = new SlowestKey();
17 System.out.println(solution.slowestKey(new int[]{12, 23, 36, 46, 62}, "spuda"));
18 }
19}The Java version uses character comparison and follows identical logic to the previous solutions. It systematically checks the each duration and updates the result accordingly.
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.
1#include <vector>
2#include <string>
using namespace std;
char slowestKeyWithArray(vector<int>& releaseTimes, string keysPressed) {
vector<int> durations(releaseTimes.size());
durations[0] = releaseTimes[0];
for (int i = 1; i < releaseTimes.size(); ++i) {
durations[i] = releaseTimes[i] - releaseTimes[i - 1];
}
int maxDuration = durations[0];
char result = keysPressed[0];
for (int i = 1; i < durations.size(); ++i) {
if (durations[i] > maxDuration || (durations[i] == maxDuration && keysPressed[i] > result)) {
maxDuration = durations[i];
result = keysPressed[i];
}
}
return result;
}Utilizes C++ vectors to handle durations, making dynamic sizing easier and clean with the STL vector operations.