681. Next Closest Time

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LeetCode

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Problem

Given a time represented in the format “HH:MM”, form the next closest time by reusing the current digits. There is no limit on how many times a digit can be reused.

You may assume the given input string is always valid. For example, “01:34”, “12:09” are all valid. “1:34”, “12:9” are all invalid.

Example 1:

Input: time = "19:34"
Output: "19:39"
Explanation: The next closest time choosing from digits 1, 9, 3, 4, is 19:39, which occurs 5 minutes later.
It is not 19:33, because this occurs 23 hours and 59 minutes later.

Example 2:

Input: time = "23:59"
Output: "22:22"
Explanation: The next closest time choosing from digits 2, 3, 5, 9, is 22:22.
It may be assumed that the returned time is next day's time since it is smaller than the input time numerically.

Constraints:

• time.length == 5
• time is a valid time in the form “HH:MM”.
• 0 <= HH < 24
• 0 <= MM < 60

Code

class Solution {
    public String nextClosestTime(String time) {
        char[] arr = time.toCharArray();
        char[] digits = new char[]{arr[0], arr[1], arr[3], arr[4]};

        Arrays.sort(digits);

        char curr = findNext(digits, arr[4], '9');
        if(arr[4] < curr) {
            arr[4] = curr;
            return new String(arr);
        }
        arr[4] = curr;

        curr = findNext(digits, arr[3], '5');
        if(arr[3] < curr) {
            arr[3] = curr;
            return new String(arr);
        }
        arr[3] = curr;

        curr = findNext(digits, arr[1], arr[0] == '2' ? '3' : '9');
        if(arr[1] < curr) {
            arr[1] = curr;
            return new String(arr);
        }
        arr[1] = curr;

        curr = findNext(digits, arr[0], '2');
        arr[0] = curr;
        return new String(arr);
    }

    private char findNext(char[] digits, char curr, char max){
        int next = -1;
        for(int i = 0; i < digits.length; i++) {
            if(digits[i] == curr) next = i;
        }

        next++;

        if(next >= digits.length || digits[next] > max) return digits[0];
        return digits[next];
    }
}