401 Binary Watch

401. Binary Watch​
1. Question
A binary watch has 4 LEDs on the top which represent thehours(0-11), and the 6 LEDs on the bottom represent theminutes(0-59).
Each LED represents a zero or one, with the least significant bit on the right.
Could not load image
For example, the above binary watch reads "3:25".
Given a non-negative integernwhich represents the number of LEDs that are currently on, return all possible times the watch could represent.
Example:
1
Input: n = 1
2
​
3
​
4
Return: ["1:00", "2:00", "4:00", "8:00", "0:01", "0:02", "0:04", "0:08", "0:16", "0:32"]
Copied!
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".
2. Implementation
(1) Backtracking
思路: 根据给定的输入num，我们生成所有可能的hours和minutes值，其中要注意排除无效的值，比如hours必须小于12，minutes必须小于60
1
public class Solution {
2
    /*
3
     * Given a num, find the combination from hours array {8, 4, 2, 1}
4
     * and minutes array {32, 16, 8, 4, 2, 1}
5
     */
6
    public List<String> readBinaryWatch(int num) {
7
        List<String> res = new ArrayList<>();
8
        int[] hours = {8, 4, 2, 1};
9
        int[] minutes = {32, 16, 8, 4, 2, 1};
10
​
11
        for (int i = 0; i <= num; i++) {
12
            // Get i elements from hours array and the remaining (num - i) from minutes
13
             List<Integer> hoursList = generateDigit(hours, i, 12);
14
             List<Integer> minutesList = generateDigit(minutes, num - i, 60);
15
             for (int hour : hoursList) {
16
                for (int min : minutesList) {
17
                    // Convert digit to correct time form
18
                    res.add(hour + ":" + (min < 10 ? "0" + min : min));
19
                }
20
             }
21
        }
22
        return res;
23
    }
24
​
25
    public List<Integer> generateDigit(int[] nums, int count, int max) {
26
        List<Integer> res = new ArrayList<>();
27
        recGenerateDigit(nums, count, 0, 0, max, res);
28
        return res;
29
    }
30
​
31
    public void recGenerateDigit(int[] nums, int count, int pos, int sum, int max, List<Integer> res) {
32
        // Exclude invalid value of hour (>= 12) and minute (>= 60)
33
        if (sum >= max) {
34
            return;
35
        }
36
​
37
        if (count == 0) {
38
            res.add(sum);
39
            return;
40
        }
41
​
42
        for (int i = pos; i < nums.length; i++) {
43
            recGenerateDigit(nums, count - 1, i + 1, sum + nums[i], max, res);
44
        }
45
    }
46
}
Copied!
3. Time & Space Complexity
Backtracking: 时间复杂度O(12 * 60) => O(1), 空间复杂度O(12 * 60) => O(1)

357 Count Numbers with Unique Digits

411 Minimum Unique Word Abbreviation

Last modified 2yr ago

Copy link

Contents

401. Binary Watch

1. Question

2. Implementation

3. Time & Space Complexity

401. Binary Watch​

1. Question

2. Implementation

3. Time & Space Complexity

401. Binary Watch