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1736. Latest Time by Replacing Hidden Digits

Description

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.

Solutions

Solution 1: Greedy

We process each digit of the string in order, following these rules:

  1. First digit: If the value of the second digit is determined and falls within the range $[4, 9]$, then the first digit can only be $1$. Otherwise, the first digit can be up to $2$.
  2. Second digit: If the value of the first digit is determined and is $2$, then the second digit can be up to $3$. Otherwise, the second digit can be up to $9$.
  3. Third digit: The third digit can be up to $5$.
  4. Fourth digit: The fourth digit can be up to $9$.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

  • class Solution {
    public String maximumTime(String time) {
        char[] t = time.toCharArray();
        if (t[0] == '?') {
            t[0] = t[1] >= '4' && t[1] <= '9' ? '1' : '2';
        }
        if (t[1] == '?') {
            t[1] = t[0] == '2' ? '3' : '9';
        }
        if (t[3] == '?') {
            t[3] = '5';
        }
        if (t[4] == '?') {
            t[4] = '9';
        }
        return new String(t);
    }
}

  • class Solution {
public:
    string maximumTime(string time) {
        if (time[0] == '?') {
            time[0] = (time[1] >= '4' && time[1] <= '9') ? '1' : '2';
        }
        if (time[1] == '?') {
            time[1] = (time[0] == '2') ? '3' : '9';
        }
        if (time[3] == '?') {
            time[3] = '5';
        }
        if (time[4] == '?') {
            time[4] = '9';
        }
        return time;
    }
};

  • class Solution:
    def maximumTime(self, time: str) -> str:
        t = list(time)
        if t[0] == '?':
            t[0] = '1' if '4' <= t[1] <= '9' else '2'
        if t[1] == '?':
            t[1] = '3' if t[0] == '2' else '9'
        if t[3] == '?':
            t[3] = '5'
        if t[4] == '?':
            t[4] = '9'
        return ''.join(t)


  • func maximumTime(time string) string {
	t := []byte(time)
	if t[0] == '?' {
		if t[1] >= '4' && t[1] <= '9' {
			t[0] = '1'
		} else {
			t[0] = '2'
		}
	}
	if t[1] == '?' {
		if t[0] == '2' {
			t[1] = '3'
		} else {
			t[1] = '9'
		}
	}
	if t[3] == '?' {
		t[3] = '5'
	}
	if t[4] == '?' {
		t[4] = '9'
	}
	return string(t)
}

  • /**
 * @param {string} time
 * @return {string}
 */
var maximumTime = function (time) {
    const t = Array.from(time);
    if (t[0] === '?') {
        t[0] = t[1] >= '4' && t[1] <= '9' ? '1' : '2';
    }
    if (t[1] === '?') {
        t[1] = t[0] == '2' ? '3' : '9';
    }
    if (t[3] === '?') {
        t[3] = '5';
    }
    if (t[4] === '?') {
        t[4] = '9';
    }
    return t.join('');
};

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