Binary Watch

A binary watch has 4 LEDs on the top which represent the hours (0-11), and the 6 LEDs on the bottom represent the minutes (0-59).

Each LED represents a zero or one, with the least significant bit on the right.

For example, the above binary watch reads "3:25".

Given a non-negative integer n which represents the number of LEDs that are currently on, return all possible times the watch could represent.

Example:

Input: n = 1 Return: ["1:00", "2:00", "4:00", "8:00", "0:01", "0:02", "0:04", "0:08", "0:16", "0:32"]

Note: The order of output does not matter. The hour must not contain a leading zero, for example "01:00" is not valid, it should be "1:00". The minute must be consist of two digits and may contain a leading zero, for example "10:2" is not valid, it should be "10:02".

Solution: Backtracking

This problem is quite similar to the combination problem, we just need to pick n elements from hours and minutes array.

The tricky part is how to combine them together. We can use the total size of both array, and iterate through both of them

• If index is less than the size of hour array, we pick from hour array
• Else we pick from the minutes array

class Solution {
private:
    vector<int> hour = {1, 2, 4, 8};
    vector<int> minute = {1, 2, 4, 8, 16, 32};

public:
    void backtrack(int start, int num, pair<int, int> time, vector<string>& ans) {
        if (num == 0) {
            ans.push_back(to_string(time.first) +  (time.second < 10 ?  ":0" : ":") + to_string(time.second));
            return;
        }

        for (int i = start; i < hour.size() + minute.size(); i++) {
            if (i < hour.size()) {    
                time.first += hour[i];
                if (time.first < 12) backtrack(i+1, num-1, time, ans);
                time.first -= hour[i];
            } else {     
                time.second += minute[i - hour.size()];
                if (time.second < 60) backtrack(i+1, num-1, time, ans);
                time.second -= minute[i - hour.size()];
            }
        }
    }

    vector<string> readBinaryWatch(int num) {
        vector<string> ans;
        backtrack(0, num, make_pair(0, 0), ans);
        return ans;
    }
};

Solution: Using bitset

    vector<string> readBinaryWatch(int num) {
        vector<string> res;
        for (int h = 0; h < 12; ++h) {
            for (int m = 0; m < 60; ++m) {
                if (bitset<10>((h << 6) + m).count() == num) {
                    res.push_back(to_string(h) + (m < 10 ? ":0" : ":") + to_string(m));
                }
            }
        }
        return res;
    }