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<int>digits; min_diff = INT_MAX; int pos = time.find(‘:‘); for(int i = 0;i<pos;i++) digits.push_back(time[i]-‘0‘); for(int i = pos+1;i<time.size();i++) digits.push_back(time[i]-‘0‘); if(isSameDigits(digits)) return time; cout << time.substr(0,pos) << endl; cout << time.substr(pos+1) << endl; curr_time = stoll(time.substr(0,pos))*60+stoll(time.substr(pos+1)); string tmp,res; unordered_set<string>set; set.insert(time); string _time; dfs(digits,_time,set,res); return res; } void dfs(vector<int>digits,string _time,unordered_set<string>&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.size();i++) dfs(digits,_time+to_string(digits[i]),set,res); } private: int curr_time; int min_diff; int calculate(string hour,string minute) { int curr = stoi(hour)*60+stoi(minute); int diff; if(curr<curr_time) diff = curr+24*60-curr_time; else diff = curr-curr_time; return diff; } bool isSameDigits(vector<int>digits) { for(int i=1;i<digits.size();i++) if(digits[i]!=digits[i-1]) return false; return true; } };