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 × 60), and the space complexity is O(1).
PythonJavaC++GoTypeScriptRust
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)
)(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
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
}
}(code-box)
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).
PythonJavaC++GoTypeScriptRust
class Solution:
def countTime(self, time: str) -> int:
def f(s: str, m: int) -> int:
cnt = 0
for i in range(m):
a = s[0] == '?' or (int(s[0]) == i // 10)
b = s[1] == '?' or (int(s[1]) == i % 10)
cnt += a and b
return cnt
return f(time[:2], 24) * f(time[3:], 60)(code-box)
class Solution {
public int countTime(String time) {
return f(time.substring(0, 2), 24) * f(time.substring(3), 60);
}
private int f(String s, int m) {
int cnt = 0;
for (int i = 0; i < m; ++i) {
boolean a = s.charAt(0) == '?' || s.charAt(0) - '0' == i / 10;
boolean b = s.charAt(1) == '?' || s.charAt(1) - '0' == i % 10;
cnt += a && b ? 1 : 0;
}
return cnt;
}
}(code-box)
class Solution {
public:
int countTime(string time) {
auto f = [](string s, int m) {
int cnt = 0;
for (int i = 0; i < m; ++i) {
bool a = s[0] == '?' || s[0] - '0' == i / 10;
bool b = s[1] == '?' || s[1] - '0' == i % 10;
cnt += a && b;
}
return cnt;
};
return f(time.substr(0, 2), 24) * f(time.substr(3, 2), 60);
}
};(code-box)
func countTime(time string) int {
f := func(s string, m int) (cnt int) {
for i := 0; i < m; i++ {
a := s[0] == '?' || int(s[0]-'0') == i/10
b := s[1] == '?' || int(s[1]-'0') == i%10
if a && b {
cnt++
}
}
return
}
return f(time[:2], 24) * f(time[3:], 60)
}(code-box)
function countTime(time: string): number {
const f = (s: string, m: number): number => {
let cnt = 0;
for (let i = 0; i < m; ++i) {
const a = s[0] === '?' || s[0] === Math.floor(i / 10).toString();
const b = s[1] === '?' || s[1] === (i % 10).toString();
if (a && b) {
++cnt;
}
}
return cnt;
};
return f(time.slice(0, 2), 24) * f(time.slice(3), 60);
}(code-box)
impl Solution {
pub fn count_time(time: String) -> i32 {
let f = |s: &str, m: usize| -> i32 {
let mut cnt = 0;
let first = s.chars().nth(0).unwrap();
let second = s.chars().nth(1).unwrap();
for i in 0..m {
let a = first == '?' || (first.to_digit(10).unwrap() as usize) == i / 10;
let b = second == '?' || (second.to_digit(10).unwrap() as usize) == i % 10;
if a && b {
cnt += 1;
}
}
cnt
};
f(&time[..2], 24) * f(&time[3..], 60)
}
}(code-box)
