LeetCode Solutions
681. Next Closest Time
Time: $O(1)$ Space: $O(1)$
class Solution {
public:
string nextClosestTime(string time) {
const set<char> digitsSet{time[0], time[1], time[3], time[4]};
string ans = time;
ans[4] = nextClosest(digitsSet, ans[4], '9');
if (time[4] < ans[4])
return ans;
ans[3] = nextClosest(digitsSet, ans[3], '5');
if (time[3] < ans[3])
return ans;
ans[1] = nextClosest(digitsSet, ans[1], ans[0] == '2' ? '3' : '9');
if (time[1] < ans[1])
return ans;
ans[0] = nextClosest(digitsSet, ans[0], '2');
return ans;
}
private:
char nextClosest(const set<char>& digitsSet, char digit, char limit) {
auto it = digitsSet.upper_bound(digit);
return it == end(digitsSet) || *it > limit ? *begin(digitsSet) : *it;
}
};
class Solution {
public String nextClosestTime(String time) {
char[] ans = time.toCharArray();
Character[] digits = {ans[0], ans[1], ans[3], ans[4]};
TreeSet<Character> digitsSet = new TreeSet<Character>(Arrays.asList(digits));
ans[4] = nextClosest(digitsSet, ans[4], '9');
if (time.charAt(4) < ans[4])
return new String(ans);
ans[3] = nextClosest(digitsSet, ans[3], '5');
if (time.charAt(3) < ans[3])
return new String(ans);
ans[1] = nextClosest(digitsSet, ans[1], ans[0] == '2' ? '3' : '9');
if (time.charAt(1) < ans[1])
return new String(ans);
ans[0] = nextClosest(digitsSet, ans[0], '2');
return new String(ans);
}
private char nextClosest(TreeSet<Character> digitsSet, char digit, char limit) {
Character next = digitsSet.higher(digit);
return next == null || next > limit ? digitsSet.first() : next;
}
}
class Solution:
def nextClosestTime(self, time: str) -> str:
ans = list(time)
digits = sorted(ans)
def nextClosest(digit: chr, limit: chr) -> chr:
next = bisect_right(digits, digit)
return digits[0] if next == 4 or digits[next] > limit else digits[next]
ans[4] = nextClosest(ans[4], '9')
if time[4] < ans[4]:
return ''.join(ans)
ans[3] = nextClosest(ans[3], '5')
if time[3] < ans[3]:
return ''.join(ans)
ans[1] = nextClosest(ans[1], '3' if ans[0] == '2' else '9')
if time[1] < ans[1]:
return ''.join(ans)
ans[0] = nextClosest(ans[0], '2')
return ''.join(ans)