Largest Time for Given Digits

Largest Time for Given Digits

Given an array of 4 digits, return the largest 24 hour time that can be made.

The smallest 24 hour time is 00:00, and the largest is 23:59.  Starting from 00:00, a time is larger if more time has elapsed since midnight.

Return the answer as a string of length 5.  If no valid time can be made, return an empty string.

 

Example 1:

Input: [1,2,3,4]
Output: "23:41"

Example 2:

Input: [5,5,5,5]
Output: ""

 

Note:

1. A.length == 4
2. 0 <= A[i] <= 9

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class Solution:

    def largestTimeFromDigits(self, A: List[int]) -> str:

        

        max_time = -1

        # enumerate all possibilities, with the permutation() func

        for h, i, j, k in itertools.permutations(A):

            hour = h*10 + i

            minute = j*10 + k

            if hour < 24 and minute < 60:

                max_time = max(max_time, hour * 60 + minute)

        

        if max_time == -1:

            return ""

        else:

            return "{:02d}:{:02d}".format(max_time // 60, max_time % 60)

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 class Solution {

public:

    string largestTimeFromDigits(vector<int>& A) {

           sort(begin(A), end(A), greater<int>());

        

  do if ((A[0] < 2 || (A[0] == 2 && A[1] < 4)) && A[2] < 6) 

      return to_string(A[0]) + to_string(A[1]) + ":" + to_string(A[2]) + to_string(A[3]);

  while (prev_permutation(begin(A), end(A)));

  return "";

    }

};

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class Solution {

public:

    string largestTimeFromDigits(vector<int>& A) {

        int max_time = -1;

        // prepare for the generation of permutations next.

        std::sort(A.begin(), A.end());

        do {

            int hour = A[0] * 10 + A[1];

            int minute = A[2] * 10 + A[3];

            if (hour < 24 && minute < 60) {

                int new_time = hour * 60 + minute;

                max_time = new_time > max_time ? new_time : max_time;

            }

        } while(next_permutation(A.begin(), A.end()));

        if (max_time == -1) {

            return "";

        } else {

            std::ostringstream strstream;

            strstream << std::setw(2) << std::setfill('0') << max_time / 60

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