LeetCode in Kotlin
869. Reordered Power of 2
Medium
You are given an integer n. We reorder the digits in any order (including the original order) such that the leading digit is not zero.
Return true if and only if we can do this so that the resulting number is a power of two.
Example 1:
Input: n = 1
Output: true
Example 2:
Input: n = 10
Output: false
Constraints:
1 <= n <= 109
Solution
import kotlin.math.pow
class Solution {
fun reorderedPowerOf2(n: Int): Boolean {
var i = 0
while (2.0.pow(i.toDouble()) < n.toLong() * 10) {
if (isValid(2.0.pow(i++.toDouble()).toInt().toString(), n.toString())) {
return true
}
}
return false
}
private fun isValid(a: String, b: String): Boolean {
val m: MutableMap<Char, Int> = HashMap()
val mTwo: MutableMap<Char, Int> = HashMap()
for (c in a.toCharArray()) {
m[c] = if (m.containsKey(c)) m[c]!! + 1 else 1
}
for (c in b.toCharArray()) {
mTwo[c] = if (mTwo.containsKey(c)) mTwo[c]!! + 1 else 1
}
for (entry in mTwo.entries.iterator()) {
if (!m.containsKey(entry.key) || entry.value != m[entry.key]) {
return false
}
}
return a[0] != '0' && m.size == mTwo.size
}
}