Latest Time by Replacing Hidden Digits - Practice Coding Problems

Latest Time by Replacing Hidden Digits - Problem

String Greedy Easy

You're given a time string in the format hh:mm, but some digits are mysteriously hidden and replaced with ? characters! 🕰️

Your mission is to maximize the time by strategically replacing each ? with the largest possible digit that still keeps the time valid. Valid times range from 00:00 to 23:59 (24-hour format).

Goal: Return the latest possible valid time by replacing all ? characters optimally.

Example: Given "2?:?0", you could make "20:50", "21:30", or "23:50" - but "23:50" is the latest valid time!

Input & Output

example_1.py — Basic Hidden Digits

$ Input: time = "2?:?0"

› Output: "23:50"

💡 Note: We need to replace two question marks. For position 1 (second hour digit), since the first hour digit is '2', the maximum valid value is '3' (making 23:xx). For position 3 (first minute digit), since the last minute digit is '0', we can use '5' to make xx:50. Final result: "23:50".

example_2.py — Multiple Possibilities

$ Input: time = "?4:5?"

› Output: "14:59"

💡 Note: For position 0 (first hour digit), since second hour digit is '4', we can use either '0' or '1' (04:xx or 14:xx), but '1' gives us a later time. For position 4 (second minute digit), we can use '9' to maximize minutes. Result: "14:59".

example_3.py — All Hidden

$ Input: time = "??:??"

› Output: "23:59"

💡 Note: When all digits are hidden, we want the latest possible time. Maximum valid hour is 23, maximum valid minutes is 59. So we get "23:59" which is the latest time in 24-hour format.

Constraints

time is in the format hh:mm
It is guaranteed that you can produce a valid time from the given string
The string length is always 5 characters
Valid times are between 00:00 and 23:59

Visualization

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Understanding the Visualization

Identify Broken Segments

The clock shows "2?:?0" - two digits are broken and need replacement

Fix Hour Display

Since first hour digit is '2', second digit can be max '3' (making 23:xx)

Fix Minute Display

Since last minute digit is '0', first minute digit can be max '5' (making xx:50)

Final Time

Clock now displays "23:50" - the latest possible valid time!

Key Takeaway

🎯 Key Insight: By using a greedy approach and understanding time format constraints, we can optimally replace each '?' with the maximum valid digit, achieving O(1) time complexity.

Asked in

A Amazon 15 G Google 8 ⊞ Microsoft 6 f Meta 4

The optimal solution uses a greedy approach that processes each position from left to right, choosing the maximum valid digit for each '?' character. Key insight: we can make locally optimal choices because earlier positions have more significant impact on the final time value. Time complexity: O(1), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Greedy Digit-by-Digit (Optimal)	O(1)	O(1)	Replace each '?' with the largest possible valid digit from left to right

Greedy Digit-by-Digit (Optimal) — Algorithm Steps

Process each character in the time string from left to right
For the first hour digit (position 0): choose '2' if second digit allows it, otherwise '9'
For the second hour digit (position 1): choose '3' if first digit is '2', otherwise '9'
For the first minute digit (position 3): choose '5' (since minutes go up to 59)
For the second minute digit (position 4): choose '9'
Return the modified time string

Visualization

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Step-by-Step Walkthrough

Analyze "2?:?0"

Start with leftmost question mark

Position 1: Choose '3'

Since first digit is '2', max second digit is '3' (23:xx ≤ 23:59)

Position 3: Choose '5'

Minutes range 00-59, so first minute digit max is '5'

Final Result: "23:50"

Optimal time achieved with greedy choices

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

char* latestTime(char* time) {
    char* result = malloc(6 * sizeof(char));
    strcpy(result, time);
    
    // Handle first hour digit
    if (result[0] == '?') {
        if (result[1] == '?' || result[1] <= '3') {
            result[0] = '2';
        } else {
            result[0] = '1';
        }
    }
    
    // Handle second hour digit
    if (result[1] == '?') {
        if (result[0] == '2') {
            result[1] = '3';
        } else {
            result[1] = '9';
        }
    }
    
    // Handle first minute digit
    if (result[3] == '?') {
        result[3] = '5';
    }
    
    // Handle second minute digit
    if (result[4] == '?') {
        result[4] = '9';
    }
    
    return result;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

We only check at most 5 positions (hh:mm), each taking constant time

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables, modifying input string in-place or creating one result string

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Greedy Digit-by-Digit (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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