LeetCode in Kotlin

1611. Minimum One Bit Operations to Make Integers Zero

Hard

Given an integer n, you must transform it into 0 using the following operations any number of times:

• Change the rightmost (0th) bit in the binary representation of n.
• Change the ith bit in the binary representation of n if the (i-1)th bit is set to 1 and the (i-2)th through 0th bits are set to 0.

Return the minimum number of operations to transform n into 0.

Example 1:

Input: n = 3

Output: 2

Explanation: The binary representation of 3 is “11”.

“11” -> “01” with the 2^nd operation since the 0^th bit is 1.

“01” -> “00” with the 1^st operation.

Example 2:

Input: n = 6

Output: 4

Explanation: The binary representation of 6 is “110”.

“110” -> “010” with the 2^nd operation since the 1^st bit is 1 and 0^th through 0^th bits are 0.

“010” -> “011” with the 1^st operation.

“011” -> “001” with the 2^nd operation since the 0^th bit is 1.

“001” -> “000” with the 1^st operation.

Constraints:

• 0 <= n <= 109

Solution

import java.util.LinkedList

@Suppress("NAME_SHADOWING")
class Solution {
    fun minimumOneBitOperations(n: Int): Int {
        return calc(calculateOneIndex(n))
    }

    private fun calc(indices: LinkedList<Int>): Int {
        if (indices.isEmpty()) {
            return 0
        }
        val index = indices.removeLast()
        return stepOfExp(index) - calc(indices)
    }

    private fun calculateOneIndex(n: Int): LinkedList<Int> {
        var n = n
        val result = LinkedList<Int>()
        var index = 1
        while (n > 0) {
            if (n % 2 == 1) {
                result.add(index)
            }
            n = n shr 1
            index++
        }
        return result
    }

    private fun stepOfExp(index: Int): Int {
        var index = index
        var result = 1
        while (index > 0) {
            result = result shl 1
            index--
        }
        return result - 1
    }
}

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