# 2437. Number of Valid Clock Times

## Description

You are given a string of length 5 called time, representing the current time on a digital clock in the format "hh:mm". The earliest possible time is "00:00" and the latest possible time is "23:59".

In the string time, the digits represented by the ? symbol are unknown, and must be replaced with a digit from 0 to 9.

Return an integer answer, the number of valid clock times that can be created by replacing every ? with a digit from 0 to 9.

Example 1:

Input: time = "?5:00"
Output: 2
Explanation: We can replace the ? with either a 0 or 1, producing "05:00" or "15:00". Note that we cannot replace it with a 2, since the time "25:00" is invalid. In total, we have two choices.


Example 2:

Input: time = "0?:0?"
Output: 100
Explanation: Each ? can be replaced by any digit from 0 to 9, so we have 100 total choices.


Example 3:

Input: time = "??:??"
Output: 1440
Explanation: There are 24 possible choices for the hours, and 60 possible choices for the minutes. In total, we have 24 * 60 = 1440 choices.


Constraints:

• time is a valid string of length 5 in the format "hh:mm".
• "00" <= hh <= "23"
• "00" <= mm <= "59"
• Some of the digits might be replaced with '?' and need to be replaced with digits from 0 to 9.

## Solutions

Solution 1: Enumeration

We can directly enumerate all times from $00:00$ to $23:59$, then judge whether each time is valid, if so, increment the answer.

After the enumeration ends, return the answer.

The time complexity is $O(24 \times 60)$, and the space complexity is $O(1)$.

Solution 2: Optimized Enumeration

We can separately enumerate hours and minutes, count how many hours and minutes meet the condition, and then multiply them together.

The time complexity is $O(24 + 60)$, and the space complexity is $O(1)$.

• class Solution {
public int countTime(String time) {
int ans = 0;
for (int h = 0; h < 24; ++h) {
for (int m = 0; m < 60; ++m) {
String s = String.format("%02d:%02d", h, m);
int ok = 1;
for (int i = 0; i < 5; ++i) {
if (s.charAt(i) != time.charAt(i) && time.charAt(i) != '?') {
ok = 0;
break;
}
}
ans += ok;
}
}
return ans;
}
}

• class Solution {
public:
int countTime(string time) {
int ans = 0;
for (int h = 0; h < 24; ++h) {
for (int m = 0; m < 60; ++m) {
char s[20];
sprintf(s, "%02d:%02d", h, m);
int ok = 1;
for (int i = 0; i < 5; ++i) {
if (s[i] != time[i] && time[i] != '?') {
ok = 0;
break;
}
}
ans += ok;
}
}
return ans;
}
};

• class Solution:
def countTime(self, time: str) -> int:
def check(s: str, t: str) -> bool:
return all(a == b or b == '?' for a, b in zip(s, t))

return sum(
check(f'{h:02d}:{m:02d}', time) for h in range(24) for m in range(60)
)


• func countTime(time string) int {
ans := 0
for h := 0; h < 24; h++ {
for m := 0; m < 60; m++ {
s := fmt.Sprintf("%02d:%02d", h, m)
ok := 1
for i := 0; i < 5; i++ {
if s[i] != time[i] && time[i] != '?' {
ok = 0
break
}
}
ans += ok
}
}
return ans
}

• function countTime(time: string): number {
let ans = 0;
for (let h = 0; h < 24; ++h) {
for (let m = 0; m < 60; ++m) {
const s = ${h}.padStart(2, '0') + ':' + ${m}.padStart(2, '0');
let ok = 1;
for (let i = 0; i < 5; ++i) {
if (s[i] !== time[i] && time[i] !== '?') {
ok = 0;
break;
}
}
ans += ok;
}
}
return ans;
}


• impl Solution {
pub fn count_time(time: String) -> i32 {
let mut ans = 0;

for i in 0..24 {
for j in 0..60 {
let mut ok = true;
let t = format!("{:02}:{:02}", i, j);

for (k, ch) in time.chars().enumerate() {
if ch != '?' && ch != t.chars().nth(k).unwrap() {
ok = false;
}
}

if ok {
ans += 1;
}
}
}

ans
}
}