Latest Time by Replacing Hidden Digits

🏠 ⬅️ ➡️

You are given a string time in the form of hh:mm, where some of the digits in the string are hidden (represented by ?).

The valid times are those inclusively between 00:00 and 23:59.

Return the latest valid time you can get from time by replacing the hidden digits.

Example 1:

Input: time = "2?:?0 " Output: "23:50 " Explanation: The latest hour beginning with the digit '2' is 23 and the latest minute ending with the digit '0' is 50.

Example 2:

Input: time = "0?:3? " Output: "09:39 "

Example 3:

Input: time = "1?:22 " Output: "19:22 "

Constraints:

  • time is in the format hh:mm.
  • It is guaranteed that you can produce a valid time from the given string.
Note: This problem is from LeetCode.
❌ Compiled
❌ Executed
❌ Correct 
program main
    implicit none

    character(len=8) :: time
    integer :: hh, mm

    time = "2?:?0"
    call solve(time, hh, mm)
    write (*, "(A, I2, A, I2)") "Latest valid time from ", time, " is ", hh, ":", mm

    time = "0?:3?"
    call solve(time, hh, mm)
    write (*, "(A, I2, A, I2)") "Latest valid time from ", time, " is ", hh, ":", mm

    time = "1?:22"
    call solve(time, hh, mm)
    write (*, "(A, I2, A, I2)") "Latest valid time from ", time, " is ", hh, ":", mm

contains

    subroutine solve(time, hh, mm)
        implicit none
        character(len=*), intent(in) :: time
        integer, intent(out) :: hh, mm

        integer :: i, j
        logical :: found

        do i = 1, 2
            if (time(i:i) == '?') then
                do j = 0, 9
                    if (j /= 0 .and. j /= 6) then
                        time(i:i) = char(ichar('0') + j)
                        found = .true.
                        exit
                    end if
                end do
                if (.not. found) then
                    stop "No solution"
                end if
            end if
        end do

        read (time, "(I2, A, I2)") hh, ":", mm
    end subroutine solve
end program main
✅ Compiled
❌ Executed
❌ Correct 
PROGRAM main

IMPLICIT NONE

CHARACTER(len=8) :: time
CHARACTER(len=2) :: hour
CHARACTER(len=2) :: minute
INTEGER :: h, m

! Read input
READ(*,*) time

! Extract hour and minute
hour = time(1:2)
minute = time(4:5)

! Convert to integers
READ(hour,*) h
READ(minute,*) m

! Check if time is valid
IF (h < 24 .AND. m < 60) THEN
    ! Print latest valid time
    WRITE(*,*) hour, minute
ELSE
    ! Print error message
    WRITE(*,*) "Invalid time"
END IF

END PROGRAM main
🌐 Data from online sources 
def maximumTime(time: str) -> str:
    time_list = list(time)
    if time_list[0] == '?':
        time_list[0] = '2' if time_list[1] <= '3' or time_list[1] == '?' else '1'
    if time_list[1] == '?':
        time_list[1] = '3' if time_list[0] == '2' else '9'
    if time_list[3] == '?':
        time_list[3] = '5'
    if time_list[4] == '?':
        time_list[4] = '9'
    return "".join(time_list)

The algorithm uses the following steps: 1. For each character in the input string, check if it's a "?". 2. If the character is "?", assign a valid hour or minute value depending on its position in the string and other characters' values. - If the first "?" is at position 0, check if the character at position 1 is less than or equal to 3 or "?". If true, assign '2' else, assign '1' to position 0. - If the first "?" is at position 1, check if the character at position 0 is '2'. If true, assign '3' else, assign '9' to position 1. - If the first "?" is at position 3, assign '5' to it, any value between 00-59 is valid. - If the first "?" is at position 4, assign '9' to it, any value between 00-59 is valid.

Finally, return the modified string.

🌐 Data from online sources 
#include <string>

std::string maximumTime(std::string time) {
    if (time[0] == '?') {
        time[0] = (time[1] <= '3' || time[1] == '?') ? '2' : '1';
    }
    if (time[1] == '?') {
        time[1] = (time[0] == '2') ? '3' : '9';
    }
    if (time[3] == '?') {
        time[3] = '5';
    }
    if (time[4] == '?') {
        time[4] = '9';
    }
    return time;
}

The algorithm uses the following steps: 1. For each character in the input string, check if it's a "?". 2. If the character is "?", assign a valid hour or minute value depending on its position in the string and other characters' values. - If the first "?" is at position 0, check if the character at position 1 is less than or equal to 3 or "?". If true, assign '2' else, assign '1' to position 0. - If the first "?" is at position 1, check if the character at position 0 is '2'. If true, assign '3' else, assign '9' to position 1. - If the first "?" is at position 3, assign '5' to it, any value between 00-59 is valid. - If the first "?" is at position 4, assign '9' to it, any value between 00-59 is valid.

Finally, return the modified string.