1736. Latest Time by Replacing Hidden Digits

Problem

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.

Solution (Java)

class Solution {
    public String maximumTime(String time) {
        StringBuilder sb = new StringBuilder();
        String[] strs = time.split(":");
        String hour = strs[0];
        String min = strs[1];
        if (hour.charAt(0) == '?') {
            if (hour.charAt(1) == '?') {
                sb.append("23");
            } else if (hour.charAt(1) > '3') {
                sb.append("1");
                sb.append(hour.charAt(1));
            } else {
                sb.append("2");
                sb.append(hour.charAt(1));
            }
        } else if (hour.charAt(0) == '0' || hour.charAt(0) == '1') {
            if (hour.charAt(1) == '?') {
                sb.append(hour.charAt(0));
                sb.append("9");
            } else {
                sb.append(hour);
            }
        } else if (hour.charAt(0) == '2') {
            if (hour.charAt(1) == '?') {
                sb.append("23");
            } else {
                sb.append(hour);
            }
        }
        sb.append(":");
        if (min.charAt(0) == '?') {
            if (min.charAt(1) == '?') {
                sb.append("59");
            } else {
                sb.append("5");
                sb.append(min.charAt(1));
            }
            return sb.toString();
        }
        sb.append(min.charAt(0));
        if (min.charAt(1) == '?') {
            sb.append("9");
        } else {
            sb.append(min.charAt(1));
        }
        return sb.toString();
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).