LeetCode: Binary Watch

LeetCode: 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.

See the picture here: https://leetcode.com/problems/binary-watch/description/ 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:

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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:

  1. The order of output does not matter.
  2. The hour must not contain a leading zero, for example “01:00” is not valid, it should be “1:00”.
  3. 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”.
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public class Solution {
public List<String> readBinaryWatch(int num) {
if (num < 0 || num > 10) return null;

List<String> result = new ArrayList<>();

for (int i = Math.max(0, num - 6); i < Math.min(num+1, 4); i++) {
Set<Integer> hours = new HashSet<>();
helper(hours, new int[]{1, 2, 4, 8}, 0, 0, i, 0);

Set<Integer> minutes = new HashSet<>();
helper(minutes, new int[]{1, 2, 4, 8, 16, 32}, 0, 0, num - i, 0);

for (int x : hours) {
if(x>11) continue;
for (int y : minutes) {
if(y>59) continue;
result.add(x + ":" + (y < 10 ? "0" + y : y));
}
}
}
Collections.sort(result);
return result;
}

public void helper(Set<Integer> result, int[] nums, int num, int start, int total, int count) {
if (count == total) {
result.add(num);
return;
}

for (int i = start; i < nums.length; i++) {
int c = nums[i];
if ((num & c) == 0) {
helper(result, nums, num | c, start + 1, total, count + 1);
}
helper(result, nums, num, start + 1, total, count);
}
}
}