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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

Solution 1: Enumerate Hours and Minutes

Thinking

Form the latest valid time from four digits. There are only \(24\times 60\) legal hour-minute pairs. Enumerate them from large to small and accept the first whose digit counts match the input.

Enumerate valid hours \(h \in [0,23]\) and minutes \(m \in [0,59]\) from large to small, and use a count array to check whether the four digits match the input. The first hit is the latest valid time.

The time complexity is \(O(1)\), and the space complexity is \(O(1)\).

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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 ''

Solution 2: Brute Force (Permutations)

Thinking

Enumerating times is constant work; we may instead permute the four indices, build an hour and a minute, and keep the maximum legal value. \(4!=24\) permutations are equally affordable.

Enumerate all permutations of the four digits, check whether they form a valid time, and keep the maximum.

The time complexity is \(O(4^3)\), and the space complexity is \(O(1)\).

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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}'

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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);
    }
}

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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);
    }
};

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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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