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949. Largest Time for Given Digits

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

• Java
• C++
• Python
• Go

• 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);
      }
  }
  

• 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);
      }
  };
  

• 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}'
  
  

• 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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