<<< Decimal to Binary | Logical Flow of Code (LFoC)

The given code represents a method to convert a decimal number to its binary representation. Here's a detailed description of the code

1. Initialize variables:

   - `n`: The decimal number to be converted to binary.

   - `i`: A counter variable used to keep track of the binary digits.

   - `a`: An array to store the binary digits (binary representation of the decimal number).

2. Perform the binary conversion using a loop:

   - The while loop runs until the decimal number `n` becomes 0.

   - Inside the loop, the remainder (`rem`) is calculated by taking the modulo 2 of the current value of `n`. This gives the rightmost bit of the binary representation of `n`.

   - The value of `n` is then updated by dividing it by 2, effectively shifting its binary representation one bit to the right, discarding the least significant bit.

   - The remainder `rem` is stored in the array `a` at index `i`, effectively storing the binary digit in reverse order (from right to left) as it's easier to calculate this way.

   - The counter variable `i` is incremented to move to the next index of the array `a`.

3. Reverse the binary representation:

   - After the while loop, the value of `i` will be one more than the actual number of binary digits in the array `a`. So, to get the correct count, we decrement `i` by 1.

4. Print the binary representation:

   - The for loop starts with `i` being the index of the last binary digit and goes backwards until reaching the first binary digit (index 0).

   - Inside the loop, each binary digit is printed, giving the final binary representation of the decimal number in the correct order.

Example:

Suppose we have the input decimal number `25`.

1. The loop iterations will be as follows:

   - `n = 25, rem = 25 % 2 = 1, n = 25 / 2 = 12, a = [1], i = 1`

   - `n = 12, rem = 12 % 2 = 0, n = 12 / 2 = 6, a = [0, 1], i = 2`

   - `n = 6, rem = 6 % 2 = 0, n = 6 / 2 = 3, a = [0, 0, 1], i = 3`

   - `n = 3, rem = 3 % 2 = 1, n = 3 / 2 = 1, a = [1, 0, 0, 1], i = 4`

   - `n = 1, rem = 1 % 2 = 1, n = 1 / 2 = 0, a = [1, 1, 0, 0, 1], i = 5`

   - `n = 0, loop ends`

2. Reverse the binary representation:

   - `i = 5 - 1 = 4`

3. Print the binary representation:

   - `a[4] = 1, a[3] = 1, a[2] = 0, a[1] = 0, a[0] = 1`

   - The binary representation of `25` is `11001`.

Thus, for the input `25`, the code will output `11001`, which is the correct binary representation of the decimal number `25`.

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