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.

class Solution {public:    string nextClosestTime(string time) {        vector
     
      digits;        min_diff = INT_MAX;        int pos = time.find(':');        for(int i = 0;i
      
       set;        set.insert(time);        string _time;        dfs(digits,_time,set,res);         return res;    }    void dfs(vector
       
        digits,string _time,unordered_set
        
         &set,string &res)    {                if(_time.size()==digits.size())        {            for(int j = 1;j<_time.size();j++)            {                string hour = _time.substr(0,j);                string min = _time.substr(j);                if(hour.size()>2 || min.size()>2) continue;                if(stoi(hour)>24 || stoi(min)>=60) continue;                if(set.find(hour+":"+min)!=set.end()) continue;                set.insert(hour+":"+min);                               int curr_diff = calculate(hour,min);                cout << "curr_diff = " << curr_diff << " for time:" << hour+":"+min << endl;                if(curr_diff < min_diff)                {                    min_diff = curr_diff;                    res = hour+":"+min;                }            }            return;        }        for(int i = 0;i
         

digits) { for(int i=1;i