Problem Statement

Given a positive integer n, write a function that returns its binary equivalent as a string. The function should not use any in-built binary conversion function.

Examples

Example 1:

Input: 2
Output: "10"
Explanation: The binary equivalent of 2 is 10.

Example 2:

Input: 7
Output: "111"
Explanation: The binary equivalent of 7 is 111.

Example 3:

Input: 18
Output: "10010"
Explanation: The binary equivalent of 18 is 10010.

Constraints:

• 1 <= n <= 10⁹

Solution

We can use a stack to efficiently create the binary representation of a given decimal number. Our algorithm will take advantage of the 'Last In, First Out' property of stacks to reverse the order of the binary digits, since the binary representation is constructed from least significant bit to most significant bit, but needs to be displayed in the opposite order. The procedure involves repeatedly dividing the decimal number by 2, pushing the remainder onto the stack, which corresponds to the binary digit. When the number is reduced to 0, the algorithm pops elements off the stack and appends to the result string until the stack is empty, thereby reversing the order of digits. The result is the binary equivalent of the input decimal number.

Here is a detailed walkthrough of the solution:

1. First, the algorithm starts by creating an empty stack. A stack is chosen because of its "Last In, First Out" property which is perfect for this type of problem where we need to reverse the order of the operations.

2. Then, it enters into a loop where the given number is repeatedly divided by 2. This is because the binary number system is base 2, and each bit represents a power of 2.

3. Inside the loop, the remainder when the number is divided by 2 (which is either 0 or 1) is pushed onto the stack. This remainder is essentially the bit in the binary representation of the number.

4. The number is then updated by integer division by 2 (in programming languages, this is usually denoted by num //= 2 or num = Math.floor(num / 2) or num /= 2). This step essentially "shifts" the number one bit to the right.

5. Steps 3 and 4 are repeated until the number becomes zero.

6. At this point, the stack contains the binary representation of the number, but in reverse order. This is because the first bit we calculated (the least significant bit, or the "rightmost" bit) is on the bottom of the stack, while the last bit we calculated (the most significant bit, or the "leftmost" bit) is on the top of the stack.

7. So, the next step is to reverse this order. This is done by popping the stack until it's empty and appending each popped bit to the result. Since a stack follows "Last In, First Out" rule, this will correctly reverse the order of the bits.

8. Finally, the algorithm returns the result, which is the binary representation of the original number.

Complexity Analysis

Time Complexity

• Converting decimal to binary: The algorithm repeatedly divides the input number by 2 to get the binary digits. The number of times the division occurs is proportional to the number of bits in the binary representation of the number. For a number num, the number of iterations is O(\log_2(num)), because the number of binary digits required to represent num is \log_2(num).

• Stack operations: Each division operation results in a binary digit being pushed onto the stack, which takes constant time, O(1). Then, we pop each binary digit from the stack to append it to the result string, which also takes O(\log_2(num)).

• Therefore, the total time complexity is O(\log_2(num)).

Overall time complexity: O(\log_2(num)).

Space Complexity

• Stack space: The stack stores each binary digit. The number of binary digits (bits) required to represent a number num is \log_2(num), so the stack will require O(\log_2(num)) space.

• StringBuilder: The StringBuilder is used to construct the binary representation, which will contain the same number of bits as the stack, i.e., O(\log_2(num)).

Overall space complexity: O(\log_2(num)).