LeetCode 0949. Largest Time for Given Digits Solution in Java, C++, Python & Go | Explanation + Code

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 == 4
• 0 <= arr[i] <= 9

Solutions

Solution 1

PythonJavaC++Go

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 ''(code-box)

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); } }(code-box)

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); } };(code-box)

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) }(code-box)

Solution 2

Python

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}'(code-box)