LeetCode in Kotlin
3309. Maximum Possible Number by Binary Concatenation
Medium
You are given an array of integers nums of size 3.
Return the maximum possible number whose binary representation can be formed by concatenating the binary representation of all elements in nums in some order.
Note that the binary representation of any number does not contain leading zeros.
Example 1:
Input: nums = [1,2,3]
Output: 30
Explanation:
Concatenate the numbers in the order [3, 1, 2] to get the result "11110", which is the binary representation of 30.
Example 2:
Input: nums = [2,8,16]
Output: 1296
Explanation:
Concatenate the numbers in the order [2, 8, 16] to get the result "10100010000", which is the binary representation of 1296.
Constraints:
nums.length == 31 <= nums[i] <= 127
Solution
class Solution {
private var result = "0"
fun maxGoodNumber(nums: IntArray): Int {
val visited = BooleanArray(nums.size)
val sb = StringBuilder()
solve(nums, visited, 0, sb)
var score = 0
var `val`: Int
for (c in result.toCharArray()) {
`val` = c.code - '0'.code
score *= 2
score += `val`
}
return score
}
private fun solve(nums: IntArray, visited: BooleanArray, pos: Int, sb: StringBuilder) {
if (pos == nums.size) {
val `val` = sb.toString()
if ((result.length == `val`.length && result.compareTo(`val`) < 0) ||
`val`.length > result.length
) {
result = `val`
}
return
}
var cur: String?
for (i in nums.indices) {
if (visited[i]) {
continue
}
visited[i] = true
cur = Integer.toBinaryString(nums[i])
sb.append(cur)
solve(nums, visited, pos + 1, sb)
sb.setLength(sb.length - cur.length)
visited[i] = false
}
}
}