# 401. Binary Watch

## Approach 1: Brute Force

`class Solution(object):    def readBinaryWatch(self, num):        """        https://leetcode.com/problems/binary-watch/description/        :type num: int        :rtype: List[str]        """        return ['%d:%02d' % (h, m)            for h in range(12) for m in range(60)            if (bin(h) + bin(m)).count('1') == num]`

Iterating through each of the hours and each of the minutes and checking for the number of ones in them to populate the list. There are 720 iterations for every input.

This solution although not the best comes in at 82.72% percentile

Time Complexity: O(1)

Space Complexity: O(1)

## Approach 2: A very very very bad solution

`class Solution(object):    def readBinaryWatch(self, num):        """        :type num: int        :rtype: List[str]        """        if num > 8:            return []    	return [["0:00"], 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 	["3:31","3:47","3:55","3:59","5:31","5:47","5:55","5:59","6:31","6:47","6:55","6:59","7:15","7:23","7:27","7:29","7:30","7:39","7:43","7:45","7:46","7:51","7:53","7:54","7:57","7:58","9:31","9:47","9:55","9:59","10:31","10:47","10:55","10:59","11:15","11:23","11:27","11:29","11:30","11:39","11:43","11:45","11:46","11:51","11:53","11:54","11:57","11:58"],["7:31","7:47","7:55","7:59","11:31","11:47","11:55","11:59"]][num]        `

This solution stores all the times in a list and just returns appropriate answer when requested

Time Complexity: O(1), fixed object size

Space Complexity: O(1), fixed object size