# 681.Next-Closest-Time

## 681. Next Closest Time

## 题目地址

<https://leetcode.com/problems/next-closest-time/>

## 题目描述

```
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: "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: "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.
```

## 代码

### Approach #1

Time Complexity: O(1). We try up to 24 \* 60 possible times until we find the correct time

Space Complexity: O(1)

```java
class Solution {
  public String nextClosestTime(String time) {
        int cur = 60 * Integer.parseInt(time.substring(0, 2));
    cur += Integer.parseInt(time.substring(3));
    Set<Integer> allowed = new HashSet();
    for (char c: time.toCharArray()) {
      if (c != ':') {
        allowed.add(c - '0');
      }
    }

    while (true) {
      cur = (cur + 1) % (24 * 60);
      int[] digits = new int[] {cur / 60 / 10, cur / 60 % 10, 
                                cur % 60 / 10, cur % 60 % 10};
      search : {
        for (int d: digits) {
          if (!allowed.contains(d))
            break search;
        }
        return String.format("%02d:%02d", cur / 60, cur % 60);
      }
    }
  }
}
```

### Approach #2 Build From Allowed Digits

```java
class Solution {
  public String nextClosestTime(String time) {
    int start = 60 * Integer.parseInt(time.substring(0, 2));
    start += Integer.parseInt(time.substring(3));
    int ans = start;
    int elapsed = 24 * 60;
    Set<Integer> allowed = new HashSet();
    for (char c: time.toCharArray()) {
      if (c != ':') {
        allowed.add(c - '0');
      }
    }

    for (int h1: allowed) {
      for (int h2: allowed) {
        if (h1 * 10 + h2 < 24) {
          for (int m1: allowed) {
            for (int m2: allowed) {
              if (m1 * 10 + m2 < 60) {
                int cur = 60 * (h1 * 10 + h2) + (m1 * 10 + m2);
                int candElapsed = Math.floorMod(cur - start, 24 * 60);
                if (0 < candElapsed && candElapsed < elapsed) {
                  ans = cur;
                  elapsed = candElapsed;
                }
              }
            }
          }
        }
      }
    }

    return String.format("%02d:%02d", ans / 60, ans % 60);
  }
}
```


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