LeetCode 401: Binary Watch Java Solution Explained

Featured Snippet Summary: To solve the LeetCode 401 Binary Watch problem efficiently in Java, iterate through all 12 possible hours (0-11) and all 60 possible minutes (0-59). For each combination, use Integer.bitCount() to count the active bits. If the total number of set bits exactly matches the input integer turnedOn, format the time as "H:MM" and add it to your results array. This achieves an optimal constant time complexity.

Understanding the Binary Watch Algorithm

A binary watch is a fascinating piece of timekeeping hardware that displays time using binary sequences. It has 4 LEDs on the top row to represent the hours (0-11), and 6 LEDs on the bottom row to represent the minutes (0-59). Each individual LED represents a single zero or one in base-2 notation, with the least significant bit located on the far right.

• For example, the below binary watch reads "4:51" because the 4-hour bit is illuminated, and the 32, 16, 2, and 1 minute bits are illuminated (32+16+2+1 = 51).

Problem Constraints and Requirements

Given an integer turnedOn which represents the total number of LEDs that are currently illuminated on the watch face (ignoring any AM/PM indicators), your task is to return all possible valid times the watch could currently represent. You may return the resulting list of times in any order.

The time format rules are strict:

• The hour component must not contain a leading zero. For example, "01:00" is entirely invalid and should be properly formatted as "1:00".
• The minute component must strictly consist of two digits and must contain a leading zero when applicable. For example, "10:2" is invalid and should be properly padded as "10:02".

Example Test Cases

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

Input: turnedOn = 9
Output: []

Optimal Java Solution Implementation

Our Java implementation utilizes a straightforward brute-force technique that is both highly readable and exceptionally fast due to the small, constant domain size (720 total iterations). By leveraging Java's built-in bit manipulation utilities, we can count the active LEDs instantaneously.

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List readBinaryWatch(int turnedOn) {
        List result = new ArrayList<>();

        // Iterate over possible hours (0 to 11)
        for (int hour = 0; hour < 12; hour++) {
            // Iterate over possible minutes (0 to 59)
            for (int minute = 0; minute < 60; minute++) {
                // Count the number of bits turned on
                int bitsOn = Integer.bitCount(hour) + Integer.bitCount(minute);
                
                // If the total bits on matches the turnedOn count
                if (bitsOn == turnedOn) {
                    // Format the time as "H:MM"
                    result.add(String.format("%d:%02d", hour, minute));
                }
            }
        }
        return result;
    }
}

Algorithm Complexity Analysis

Because the nested loops always run exactly 12 * 60 = 720 times regardless of the input value, the time complexity is strictly O(1) constant time. The space complexity is also O(1) constant space, excluding the memory required to store the output results. This makes our bit-counting approach the gold standard for this specific problem compared to recursive backtracking solutions.

Frequently Asked Questions

Here are some common queries developers have when trying to implement a robust solution for the Binary Watch problem:

What is the best way to solve the Binary Watch problem in Java?

The most efficient way to solve the Binary Watch problem is to iterate through all possible hours (0-11) and minutes (0-59), and count the number of set bits (1s) in their binary representation. If the sum of set bits equals the given turnedOn value, it is a valid time.

Why do we iterate through all possible times instead of combinations?

Iterating through all 720 possible times (12 hours * 60 minutes) and counting the set bits using Integer.bitCount is much simpler and less prone to errors than generating combinations of bits, while maintaining an excellent constant time complexity of O(1).