LeetCode in Kotlin
3133. Minimum Array End
Medium
You are given two integers n and x. You have to construct an array of positive integers nums of size n where for every 0 <= i < n - 1, nums[i + 1] is greater than nums[i], and the result of the bitwise AND operation between all elements of nums is x.
Return the minimum possible value of nums[n - 1].
Example 1:
Input: n = 3, x = 4
Output: 6
Explanation:
nums can be [4,5,6] and its last element is 6.
Example 2:
Input: n = 2, x = 7
Output: 15
Explanation:
nums can be [7,15] and its last element is 15.
Constraints:
1 <= n, x <= 108
Solution
@Suppress("NAME_SHADOWING")
class Solution {
fun minEnd(n: Int, x: Int): Long {
var n = n
n -= 1
val xb = IntArray(64)
val nb = IntArray(64)
for (i in 0..31) {
xb[i] = (x shr i) and 1
nb[i] = (n shr i) and 1
}
var i = 0
var j = 0
while (i < 64) {
if (xb[i] != 1) {
xb[i] = nb[j++]
}
i++
}
var ans: Long = 0
var p: Long = 1
i = 0
while (i < 64) {
ans += (xb[i]) * p
p *= 2
i++
}
return ans
}
}