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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.

  • 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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