package com.kotlinground.bitmanipulation.countbits

/**
 * To understand the solution, we remember the following two points from math:
 *
 * All whole numbers can be represented by 2N (even) and 2N+1 (odd).
 * For a given binary number, multiplying by 2 is the same as adding a zero at the end (just as we just add a zero when
 * multiplying by ten in base 10).
 * Since multiplying by 2 just adds a zero, then any number and its double will have the same number of 1's. Likewise,
 * doubling a number and adding one will increase the count by exactly 1. Or:
 *
 * countBits(N) = countBits(2N)
 * countBits(N)+1 = countBits(2N+1)
 *
 * Thus, we can see that any number will have the same bit count as half that number, with an extra one if it's an odd
 * number. We iterate through the range of numbers and calculate each bit count successively in this manner:
 *
 *   for (i in 0 until num) {
 *      counter[i] = counter[i // 2] + i % 2
 *   }
 *
 * With a few modifications:
 *
 * - Define the base case of counter[0] = 0, and start at 1.
 * - We need to include num, so use num+1 as the bound on the range.
 * - Bit-shifting 1 has the same effect as dividing by 2, and is faster, so replace i // 2 with i >> 1.
 * - We can choose to either initiate our list with counter = [0] * (num+1) or just counter = [0] and then append to it
 * (which has O(1)). It's a little faster to initiate it with zeroes and then access it rather than appending each time,
 * but I've chosen the latter for better readability.
 * - Some solutions use i & 1 to determine the parity of i. While this accomplishes the same purpose as i % 2 and keeps
 * with the bitwise-operator theme, it is faster to calculate the modulus.
 *
 * Time and Space Complexity
 * Time: O(n) - We iterate through the range of numbers once.
 * Space: O(n) - We use a num-sized array.
 */
fun countBits(n: Int): IntArray {
    val counter = IntArray(n + 1) { 0 }
    for (i in 1 until n + 1) {
        counter[i] = counter[i.shr(1)] + i % 2
    }
    return counter
}

package com.kotlinground.bitmanipulation.countbits

/**
 * To understand the solution, we remember the following two points from math:
 *
 * All whole numbers can be represented by 2N (even) and 2N+1 (odd).
 * For a given binary number, multiplying by 2 is the same as adding a zero at the end (just as we just add a zero when
 * multiplying by ten in base 10).
 * Since multiplying by 2 just adds a zero, then any number and its double will have the same number of 1's. Likewise,
 * doubling a number and adding one will increase the count by exactly 1. Or:
 *
 * countBits(N) = countBits(2N)
 * countBits(N)+1 = countBits(2N+1)
 *
 * Thus, we can see that any number will have the same bit count as half that number, with an extra one if it's an odd
 * number. We iterate through the range of numbers and calculate each bit count successively in this manner:
 *
 *   for (i in 0 until num) {
 *      counter[i] = counter[i // 2] + i % 2
 *   }
 *
 * With a few modifications:
 *
 * - Define the base case of counter[0] = 0, and start at 1.
 * - We need to include num, so use num+1 as the bound on the range.
 * - Bit-shifting 1 has the same effect as dividing by 2, and is faster, so replace i // 2 with i >> 1.
 * - We can choose to either initiate our list with counter = [0] * (num+1) or just counter = [0] and then append to it
 * (which has O(1)). It's a little faster to initiate it with zeroes and then access it rather than appending each time,
 * but I've chosen the latter for better readability.
 * - Some solutions use i & 1 to determine the parity of i. While this accomplishes the same purpose as i % 2 and keeps
 * with the bitwise-operator theme, it is faster to calculate the modulus.
 *
 * Time and Space Complexity
 * Time: O(n) - We iterate through the range of numbers once.
 * Space: O(n) - We use a num-sized array.
 */
fun countBits(n: Int): IntArray {
    val counter = IntArray(n + 1) { 0 }
    for (i in 1 until n + 1) {
        counter[i] = counter[i.shr(1)] + i % 2
    }
    return counter
}
