Description#
Given an array arr of 4 digits, find the latest 24-hour time that can be made using each digit exactly once.
24-hour times are formatted as "HH:MM", where HH is between 00 and 23, and MM is between 00 and 59. The earliest 24-hour time is 00:00, and the latest is 23:59.
Return the latest 24-hour time in "HH:MM" format. If no valid time can be made, return an empty string.
Example 1:
Input: arr = [1,2,3,4]
Output: "23:41"
Explanation: The valid 24-hour times are "12:34", "12:43", "13:24", "13:42", "14:23", "14:32", "21:34", "21:43", "23:14", and "23:41". Of these times, "23:41" is the latest.
Example 2:
Input: arr = [5,5,5,5]
Output: ""
Explanation: There are no valid 24-hour times as "55:55" is not valid.
Constraints:
arr.length == 40 <= arr[i] <= 9
Solutions#
Solution 1#
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| class Solution:
def largestTimeFromDigits(self, arr: List[int]) -> str:
cnt = [0] * 10
for v in arr:
cnt[v] += 1
for h in range(23, -1, -1):
for m in range(59, -1, -1):
t = [0] * 10
t[h // 10] += 1
t[h % 10] += 1
t[m // 10] += 1
t[m % 10] += 1
if cnt == t:
return f'{h:02}:{m:02}'
return ''
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| class Solution {
public String largestTimeFromDigits(int[] arr) {
int ans = -1;
for (int i = 0; i < 4; ++i) {
for (int j = 0; j < 4; ++j) {
for (int k = 0; k < 4; ++k) {
if (i != j && j != k && i != k) {
int h = arr[i] * 10 + arr[j];
int m = arr[k] * 10 + arr[6 - i - j - k];
if (h < 24 && m < 60) {
ans = Math.max(ans, h * 60 + m);
}
}
}
}
}
return ans < 0 ? "" : String.format("%02d:%02d", ans / 60, ans % 60);
}
}
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| class Solution {
public:
string largestTimeFromDigits(vector<int>& arr) {
int ans = -1;
for (int i = 0; i < 4; ++i) {
for (int j = 0; j < 4; ++j) {
for (int k = 0; k < 4; ++k) {
if (i != j && j != k && i != k) {
int h = arr[i] * 10 + arr[j];
int m = arr[k] * 10 + arr[6 - i - j - k];
if (h < 24 && m < 60) {
ans = max(ans, h * 60 + m);
}
}
}
}
}
if (ans < 0) return "";
int h = ans / 60, m = ans % 60;
return to_string(h / 10) + to_string(h % 10) + ":" + to_string(m / 10) + to_string(m % 10);
}
};
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| func largestTimeFromDigits(arr []int) string {
ans := -1
for i := 0; i < 4; i++ {
for j := 0; j < 4; j++ {
for k := 0; k < 4; k++ {
if i != j && j != k && i != k {
h := arr[i]*10 + arr[j]
m := arr[k]*10 + arr[6-i-j-k]
if h < 24 && m < 60 {
ans = max(ans, h*60+m)
}
}
}
}
}
if ans < 0 {
return ""
}
return fmt.Sprintf("%02d:%02d", ans/60, ans%60)
}
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Solution 2#
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| class Solution:
def largestTimeFromDigits(self, arr: List[int]) -> str:
ans = -1
for i in range(4):
for j in range(4):
for k in range(4):
if i != j and i != k and j != k:
h = arr[i] * 10 + arr[j]
m = arr[k] * 10 + arr[6 - i - j - k]
if h < 24 and m < 60:
ans = max(ans, h * 60 + m)
return '' if ans < 0 else f'{ans // 60:02}:{ans % 60:02}'
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