LeetCode in Kotlin

2571. Minimum Operations to Reduce an Integer to 0

Medium

You are given a positive integer n, you can do the following operation any number of times:

• Add or subtract a power of 2 from n.

Return the minimum number of operations to make n equal to 0.

A number x is power of 2 if x == 2i where i >= 0.

Example 1:

Input: n = 39

Output: 3

Explanation: We can do the following operations:

• Add 2⁰ = 1 to n, so now n = 40.

• Subtract 2³ = 8 from n, so now n = 32.

• Subtract 2⁵ = 32 from n, so now n = 0. It can be shown that 3 is the minimum number of operations we need to make n equal to 0.

Example 2:

Input: n = 54

Output: 3

Explanation: We can do the following operations:

• Add 2¹ = 2 to n, so now n = 56.

• Add 2³ = 8 to n, so now n = 64.

• Subtract 2⁶ = 64 from n, so now n = 0. So the minimum number of operations is 3.

Constraints:

• 1 <= n <= 105

Solution

@Suppress("NAME_SHADOWING")
class Solution {
    fun minOperations(n: Int): Int {
        var n = n
        var count = 0
        while (n > 0) {
            val x = kotlin.math.ln(n.toDouble()) / kotlin.math.ln(2.0)
            if (x % 1.0 < 0.5) {
                n = kotlin.math.abs(n - Math.pow(2.0, x.toInt().toDouble()).toInt())
                count++
            } else {
                n = kotlin.math.abs(n - Math.pow(2.0, (x.toInt() + 1).toDouble()).toInt())
                count++
            }
        }
        return count
    }
}

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