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Formatted question description: https://leetcode.ca/all/949.html

949. Largest Time for Given Digits (Easy)

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

Related Topics:
Math

Solution 1.

• Java
• C++
• Python
• Go

• class Solution {
      public String largestTimeFromDigits(int[] A) {
          int curHour = -1, curMinute = -1;
          for (int i = 0; i < 4; i++) {
              for (int j = 0; j < 4; j++) {
                  if (j == i)
                      continue;
                  for (int k = 0; k < 4; k++) {
                      if (k == i || k == j)
                          continue;
                      int l = 6 - i - j - k;
                      int hour = A[i] * 10 + A[j], minute = A[k] * 10 + A[l];
                      if (hour <= 23 && minute <= 59) {
                          if (hour > curHour || hour == curHour && minute > curMinute) {
                              curHour = hour;
                              curMinute = minute;
                          }
                      }
                  }
              }
          }
          return curHour >= 0 ? String.format("%02d:%02d", curHour, curMinute) : "";
      }
  }
  
  ############
  
  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);
      }
  }
  

• // OJ: https://leetcode.com/problems/largest-time-for-given-digits/
  // Time: O(1)
  // Space: O(1)
  class Solution {
  private:
      string ans;
      void permute(string &s, int start) {
          if (start == s.size()) {
              ans = max(ans, s);
              return;
          }
          for (int i = start; i < s.size(); ++i) {
              if ((start == 0 && s[i] > '2')
                 || (start == 1 && s[start - 1] == '2' && s[i] > '3')
                 || (start == 2 && s[i] > '5')) continue;
              swap(s[start], s[i]);
              permute(s, start + 1);
              swap(s[start], s[i]);
          }
      }
  public:
      string largestTimeFromDigits(vector<int>& A) {
          string s;
          for (int i : A) s.push_back('0' + i);
          permute(s, 0);
          if (ans.size()) ans.insert(2, ":");
          return ans;
      }
  };
  

• 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}'
  
  ############
  
  class Solution(object):
      def largestTimeFromDigits(self, A):
          """
          :type A: List[int]
          :rtype: str
          """
          A.sort()
          for h in range(23, -1, -1):
              for m in range(59, -1, -1):
                  t = [h / 10, h % 10, m / 10, m % 10]
                  ts = sorted(t)
                  if ts == A:
                      return str(t[0]) + str(t[1]) + ":" + str(t[2]) + str(t[3])
          return ""
  

• 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)
  }
  
  func max(a, b int) int {
  	if a > b {
  		return a
  	}
  	return b
  }
  

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