Slowest Key

🏠 ⬅️ ➡️

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.
Note: This problem is from LeetCode.
✅ Compiled
❌ Executed
❌ Correct 
program main
    implicit none

    integer :: i, j, n
    character(len=10) :: keysPressed
    integer, dimension(:), allocatable :: releaseTimes
    integer :: maxDuration, currentDuration
    character(len=10) :: maxKey

    do while (.true.)
        read(*,*) n
        if (n == 0) exit

        allocate(releaseTimes(n))
        read(*,*) keysPressed
        read(*,*) releaseTimes

        maxDuration = 0
        maxKey = ""
        do i = 1, n
            currentDuration = releaseTimes(i) - releaseTimes(i - 1)
            if (currentDuration > maxDuration) then
                maxDuration = currentDuration
                maxKey = keysPressed(i:i)
            end if
        end do

        write(*,'(A)') maxKey

        deallocate(releaseTimes)
    end do
end program main
✅ Compiled
❌ Executed
❌ Correct 
program longest_keypress
      implicit none

      integer, parameter :: MAX_N = 1000
      integer :: n
      integer :: releaseTimes(MAX_N)
      character :: keysPressed(MAX_N)
      integer :: i
      integer :: longestDuration
      character :: longestKey

      read *, n
      read *, releaseTimes
      read *, keysPressed

      longestDuration = 0
      longestKey = ' '

      do i = 1, n
          if (releaseTimes(i) - releaseTimes(i - 1) > longestDuration) then
              longestDuration = releaseTimes(i) - releaseTimes(i - 1)
              longestKey = keysPressed(i)
          end if
      end do

      write (*, '(a)') longestKey

      end program longest_keypress
🌐 Data from online sources 
def minInteger(num: str, k: int) -> str:
    if k == 0:
        return num
    num = list(num)
    n = len(num)
    for i in range(n):
        idx = i
        for j in range(i + 1, n):
            if j - i > k:
                break
            if num[j] < num[idx]:
                idx = j
        min_digit = num[idx]
        for j in range(idx, i, -1):
            num[j] = num[j - 1]
        num[i] = min_digit
        k -= idx - i
    return ''.join(num)
  1. If k is 0, return the input num directly as there's no swap operation allowed.
  2. Convert the input num to an array or a list in each language.
  3. Iterate through each digit of the number (outer loop).
  4. For each digit, find the minimum digit within the range allowed by k (inner loop).
  5. Swap the found minimum digit to the current digit position (updating the array or list by shifting digits), and update k value accordingly.
  6. Convert the updated array or list back to string and return.
🌐 Data from online sources 
#include <string>
using namespace std;

string minInteger(string num, int k) {
    if (k == 0) return num;
    int n = num.size();
    for (int i = 0; i < n; ++i) {
        int idx = i;
        for (int j = i + 1; j < n && j - i <= k; ++j) {
            if (num[j] < num[idx]) {
                idx = j;
            }
        }
        char minDigit = num[idx];
        for (int j = idx; j > i; --j) {
            num[j] = num[j - 1];
        }
        num[i] = minDigit;
        k -= idx - i;
    }
    return num;
}
  1. If k is 0, return the input num directly as there's no swap operation allowed.
  2. Convert the input num to an array or a list in each language.
  3. Iterate through each digit of the number (outer loop).
  4. For each digit, find the minimum digit within the range allowed by k (inner loop).
  5. Swap the found minimum digit to the current digit position (updating the array or list by shifting digits), and update k value accordingly.
  6. Convert the updated array or list back to string and return.