Problem Description

The "Encode Number" problem asks you to encode a non-negative integer num using the following rules:

• Return a string that represents the encoded form of num.
• The encoding is based on the following pattern: given a number num, encode it as the binary representation of num + 1, but without the leading '1'.
• If num is zero, the encoded string should be an empty string.

For example:

• If num = 23, num + 1 = 24, binary is '11000'. Remove the first '1' to get '1000'.
• If num = 0, encoded string is '' (empty string).

Key constraints:

• Input: a single non-negative integer num
• Output: a string as described above

Thought Process

At first glance, you might think the problem involves some complex encoding scheme, but the rules are actually straightforward once you focus on the transformation steps:

• The main step is to convert num + 1 to its binary representation.
• Then, remove the leading '1' from this binary string.
• For example, if num = 0, then num + 1 = 1, binary is '1', and removing the only '1' leaves an empty string.
• For other numbers, the length of the encoded string is always one less than the length of num + 1 in binary.
• This is much simpler than trying to simulate any kind of recursive or iterative encoding process.

The main challenge is to correctly format the binary string and handle the edge case where num = 0.

Solution Approach

Let's break down the steps for encoding the number, and see how we can implement this efficiently:

1. Check for the Edge Case: If num == 0, immediately return an empty string. This avoids unnecessary computation.
2. Add One: Compute num + 1. This is because the encoding is based on the binary representation of num + 1.
3. Convert to Binary: Convert num + 1 to its binary representation as a string. In most languages, this is a built-in function.
4. Remove the Leading '1': The first character of the binary string for any positive number is always '1'. Remove this character to get the encoded string.
5. Return the Result: Output the final string.

This approach is efficient because:

• We only use basic arithmetic and string operations.
• There are no loops or recursion.
• Binary conversion is fast and built-in in most languages.

Example Walkthrough

Let's walk through the encoding process for num = 23:

1. Input: num = 23
2. Add One: num + 1 = 24
3. Binary Representation: 24 in binary is '11000'
4. Remove Leading '1': Remove the first character to get '1000'
5. Output: '1000'

Another example, num = 0:

• Input: num = 0
• Add One: num + 1 = 1
• Binary Representation: 1 in binary is '1'
• Remove Leading '1': Removing the only character yields an empty string
• Output: ''

Time and Space Complexity

Brute-force Approach:

• If you tried to simulate some complex encoding process, you might use recursion or repeated string operations, potentially resulting in O(log n) time and space due to the number of binary digits.

• Time Complexity: O(log n), because converting a number to binary takes time proportional to the number of bits in num + 1.
• Space Complexity: O(log n), due to the length of the binary string generated.

Both time and space are optimal for this problem because you need to output all bits except the leading one.

Summary

The "Encode Number" problem is elegantly solved by leveraging the binary representation of num + 1 and removing the leading '1'. This approach avoids unnecessary complexity and edge cases, leading to an efficient and simple solution. Key insights include recognizing the pattern in the encoding process and using basic string manipulation to achieve the desired result.

Code Implementation

class Solution:
    def encode(self, num: int) -> str:
        if num == 0:
            return ""
        binary = bin(num + 1)[2:]
        return binary[1:]
      

class Solution {
public:
    string encode(int num) {
        if (num == 0) return "";
        string binary = "";
        int n = num + 1;
        while (n > 0) {
            binary = char('0' + (n % 2)) + binary;
            n /= 2;
        }
        // Remove the leading '1'
        return binary.substr(1);
    }
};
      

class Solution {
    public String encode(int num) {
        if (num == 0) return "";
        String binary = Integer.toBinaryString(num + 1);
        return binary.substring(1);
    }
}
      

var encode = function(num) {
    if (num === 0) return "";
    let binary = (num + 1).toString(2);
    return binary.slice(1);
};