Number of Valid Clock Times - Practice Coding Problems

Number of Valid Clock Times - Problem

String Enumeration Easy

Imagine you're looking at a digital clock display, but some of the digits are broken and show up as question marks (?). Your task is to figure out how many different valid times this broken clock could be displaying!

You're given a string time of length 5 in the format "hh:mm", where:

The first two characters represent hours (00-23)
The third character is always a colon ':'
The last two characters represent minutes (00-59)
Some digits are replaced with ? symbols

Your goal: Count how many valid 24-hour format times can be formed by replacing each ? with any digit from 0 to 9.

Examples:

"?5:00" → Only "05:00", "15:00" are valid (not "25:00")
"0?:??" → Hours can be 00-09, minutes can be 00-59

Input & Output

example_1.py — Simple hour wildcard

$ Input: time = "?5:00"

› Output: 2

💡 Note: The first digit can be 0 (making 05:00) or 1 (making 15:00). It cannot be 2 or higher because 25:00 is not a valid time in 24-hour format.

example_2.py — Multiple wildcards

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

› Output: 600

💡 Note: First digit is fixed as 0, second digit can be 0-9 (giving hours 00-09), and minutes can be 00-59. So we have 10 × 60 = 600 valid combinations.

example_3.py — Edge case with constraints

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

› Output: 1440

💡 Note: All positions are wildcards. Hours can be 00-23 (24 possibilities) and minutes can be 00-59 (60 possibilities). Total: 24 × 60 = 1440 valid times.

Visualization

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Understanding the Visualization

Analyze first hour digit

If ?, can be 0,1,2. If second hour digit is 0-3, first can be 0-2. If second is 4-9, first can only be 0-1.

Analyze second hour digit

If first digit is 0-1, second can be 0-9. If first digit is 2, second can only be 0-3.

Analyze minute digits

Both minute digits can be 0-9 independently (00-59 are all valid).

Multiply possibilities

Total valid times = (valid first hour) × (valid second hour) × (valid first minute) × (valid second minute)

Key Takeaway

🎯 Key Insight: Instead of generating all combinations, analyze constraints for each position independently and multiply the valid counts!

Time & Space Complexity

Time Complexity

⏱️

O(10^k)

Where k is the number of question marks. We try all 10^k combinations.

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for validation

✓ Linear Space

Constraints

time.length == 5
time is in the format "hh:mm"
0 ≤ hh ≤ 23
0 ≤ mm ≤ 59
Some digits in time are replaced with ? symbol

Asked in

G Google 35 a Amazon 28 f Meta 22 ⊞ Microsoft 18

The optimal approach is to analyze each position independently based on clock constraints. Count valid digits for each position and multiply the counts together. This gives us O(1) time complexity instead of exponential brute force.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Combinations)	O(10^k)	O(1)	Generate all possible digit combinations and validate each one

Brute Force (Try All Combinations) — Algorithm Steps

Generate all possible combinations by replacing each '?' with digits 0-9
For each combination, check if hours are between 00-23
Check if minutes are between 00-59
Count valid combinations

Visualization

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Step-by-Step Walkthrough

Start with input

Begin with the time string containing question marks

Replace first ?

Try digits 0-9 for the first question mark

Recurse for next ?

For each valid first digit, try all possibilities for the next question mark

Validate complete time

When all ? are replaced, check if the time is valid

Code -

solution.c — C

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

bool isValidTime(const char* t) {
    int hours = (t[0] - '0') * 10 + (t[1] - '0');
    int minutes = (t[3] - '0') * 10 + (t[4] - '0');
    return hours >= 0 && hours <= 23 && minutes >= 0 && minutes <= 59;
}

int generateTimes(char* t, int pos, int len) {
    if (pos == len) {
        return isValidTime(t) ? 1 : 0;
    }
    
    if (t[pos] != '?') {
        return generateTimes(t, pos + 1, len);
    }
    
    int count = 0;
    for (int digit = 0; digit <= 9; digit++) {
        t[pos] = '0' + digit;
        count += generateTimes(t, pos + 1, len);
    }
    t[pos] = '?';
    return count;
}

int numPointsInStringOfClockTimes(char* time) {
    int len = strlen(time);
    char* copy = malloc(len + 1);
    strcpy(copy, time);
    int result = generateTimes(copy, 0, len);
    free(copy);
    return result;
}

int main() {
    char time[10];
    fgets(time, sizeof(time), stdin);
    
    // Remove quotes and newline
    char* start = strchr(time, '"');
    if (start) start++;
    else start = time;
    char* end = strrchr(start, '"');
    if (end) *end = '\0';
    char* newline = strchr(start, '\n');
    if (newline) *newline = '\0';
    
    printf("%d\n", numPointsInStringOfClockTimes(start));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(10^k)

Where k is the number of question marks. We try all 10^k combinations.

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for validation

✓ Linear Space

Constraints

time.length == 5
time is in the format "hh:mm"
0 ≤ hh ≤ 23
0 ≤ mm ≤ 59
Some digits in time are replaced with ? symbol

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Try All Combinations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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