LeetCode in Kotlin
3314. Construct the Minimum Bitwise Array I
Easy
You are given an array nums consisting of n prime integers.
You need to construct an array ans of length n, such that, for each index i, the bitwise OR of ans[i] and ans[i] + 1 is equal to nums[i], i.e. ans[i] OR (ans[i] + 1) == nums[i].
Additionally, you must minimize each value of ans[i] in the resulting array.
If it is not possible to find such a value for ans[i] that satisfies the condition, then set ans[i] = -1.
Example 1:
Input: nums = [2,3,5,7]
Output: [-1,1,4,3]
Explanation:
- For
i = 0, as there is no value forans[0]that satisfiesans[0] OR (ans[0] + 1) = 2, soans[0] = -1. - For
i = 1, the smallestans[1]that satisfiesans[1] OR (ans[1] + 1) = 3is1, because1 OR (1 + 1) = 3. - For
i = 2, the smallestans[2]that satisfiesans[2] OR (ans[2] + 1) = 5is4, because4 OR (4 + 1) = 5. - For
i = 3, the smallestans[3]that satisfiesans[3] OR (ans[3] + 1) = 7is3, because3 OR (3 + 1) = 7.
Example 2:
Input: nums = [11,13,31]
Output: [9,12,15]
Explanation:
- For
i = 0, the smallestans[0]that satisfiesans[0] OR (ans[0] + 1) = 11is9, because9 OR (9 + 1) = 11. - For
i = 1, the smallestans[1]that satisfiesans[1] OR (ans[1] + 1) = 13is12, because12 OR (12 + 1) = 13. - For
i = 2, the smallestans[2]that satisfiesans[2] OR (ans[2] + 1) = 31is15, because15 OR (15 + 1) = 31.
Constraints:
1 <= nums.length <= 1002 <= nums[i] <= 1000nums[i]is a prime number.
Solution
class Solution {
fun minBitwiseArray(nums: List<Int>): IntArray {
val l = nums.size
val r = IntArray(l)
for (i in 0 until l) {
r[i] = check(nums[i])
}
return r
}
private fun check(v: Int): Int {
if (v % 2 == 0) {
return -1
}
for (j in 1 until v) {
if ((j or (j + 1)) == v) {
return j
}
}
return -1
}
}