# 1629. Slowest Key

### 題目敘述

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.

**Hint 1:**

Get for each press its key and amount of time taken.

**Hint 2:**

Iterate on the presses, maintaining the answer so far.

**Hint 3:**

The current press will change the answer if and only if its amount of time taken is longer than that of the previous answer, or they are equal but the key is larger than that of the previous answer.

#### 題目翻譯

給一個字串 `keysPressed`

和一個排序過的佇列 `releaseTimes`

。其中相同的 `keysPressed`

的字母會重複。相同索引下，releaseTimes 的值會表示當前執行時間。

### 解法解析

這題的解法，是使用基本的 Dyanmic programming 的方式。每一次迴圈裡面去計算最慢的 key，有點類似 JavaScript 中 reduce 的功能。
也因為只有用一個迴圈處理，所以 Time complexity 只有 `O(n)`

，而只要用一個變數儲存目前的最慢的 key，所以 Space complexity 只有 `O(1)`

#### 程式範例

##### Python

```
class Solution:
def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:
n = len(releaseTimes)
max_diff = releaseTimes[0]
result = keysPressed[0]
for i in range(1, n):
diff = releaseTimes[i] - releaseTimes[i-1]
if diff > max_diff or (diff == max_diff and keysPressed[i] > result):
max_diff = diff
result = keysPressed[i]
return result
```

##### JavaScript

```
/**
* @param {number[]} releaseTimes
* @param {string} keysPressed
* @return {character}
*/
var slowestKey = function (releaseTimes, keysPressed) {
const n = releaseTimes.length;
let maxDiff = releaseTimes[0];
let result = keysPressed[0];
for (let i = 1; i < n; i++) {
const diff = releaseTimes[i] - releaseTimes[i - 1];
if (
diff > maxDiff ||
(diff === maxDiff && keysPressed[i] > result)
) {
maxDiff = diff;
result = keysPressed[i];
}
}
return result;
};
```

##### Go

```
func slowestKey(releaseTimes []int, keysPressed string) byte {
n := len(releaseTimes)
result := keysPressed[0]
maxDiff := releaseTimes[0]
for i := 1; i < n; i++ {
diff := releaseTimes[i] - releaseTimes[i-1]
if diff > maxDiff || (diff == maxDiff && keysPressed[i] > result) {
maxDiff = diff
result = keysPressed[i]
}
}
return result
}
```

##### Swift

```swiftclass Solution {
func slowestKey(* releaseTimes: [Int], * keysPressed: String) -> Character {
let keysPressed = Array(keysPressed)
var result = Character("?"), maxDiff = Int.min, lastTime = 0

```
for i in 0..<keysPressed.count {
let diff = releaseTimes[i] - lastTime
lastTime = releaseTimes[i]
if diff > maxDiff || diff == maxDiff && keysPressed[i] > result {
maxDiff = diff
result = keysPressed[i]
}
}
return result
}
```

}

```
##### Kotlin
```kotlin
class Solution {
fun slowestKey(releaseTimes: IntArray, keysPressed: String): Char {
var result = keysPressed[0]
var maxDiff = releaseTimes[0]
var index = 1
while (index < keysPressed.length) {
val currentSymbol = keysPressed[index]
val diff = releaseTimes[index] - releaseTimes[index - 1]
if (diff > maxDiff || (diff == maxDiff && currentSymbol > result)) {
maxDiff = diff
result = currentSymbol
}
index++
}
return result
}
}
```