Hard
You are given an integer n
and an array of unique integers blacklist
. Design an algorithm to pick a random integer in the range [0, n - 1]
that is not in blacklist
. Any integer that is in the mentioned range and not in blacklist
should be equally likely to be returned.
Optimize your algorithm such that it minimizes the number of calls to the built-in random function of your language.
Implement the Solution
class:
Solution(int n, int[] blacklist)
Initializes the object with the integer n
and the blacklisted integers blacklist
.int pick()
Returns a random integer in the range [0, n - 1]
and not in blacklist
.Example 1:
Input
[“Solution”, “pick”, “pick”, “pick”, “pick”, “pick”, “pick”, “pick”]
[[7, [2, 3, 5]], [], [], [], [], [], [], []]
Output: [null, 0, 4, 1, 6, 1, 0, 4]
Explanation:
Solution solution = new Solution(7, [2, 3, 5]);
solution.pick(); // return 0, any integer from [0,1,4,6] should be ok. Note that for every call of pick,
// 0, 1, 4, and 6 must be equally likely to be returned (i.e., with probability 1/4).
solution.pick(); // return 4
solution.pick(); // return 1
solution.pick(); // return 6
solution.pick(); // return 1
solution.pick(); // return 0
solution.pick(); // return 4
Constraints:
1 <= n <= 109
0 <= blacklist.length <- min(105, n - 1)
0 <= blacklist[i] < n
blacklist
are unique.2 * 104
calls will be made to pick
.import kotlin.random.Random
class Solution(n: Int, blacklist: IntArray) {
private val map: MutableMap<Int, Int>
private val upperLimit: Int
init {
map = HashMap()
upperLimit = n - blacklist.size
for (`val` in blacklist) {
map[`val`] = -1
}
var j = n - 1
for (`val` in blacklist) {
if (`val` < upperLimit) {
while (map.containsKey(j)) {
j--
}
map[`val`] = j
j--
}
}
}
fun pick(): Int {
val `val`: Int = Random.nextInt(upperLimit)
return if (map.containsKey(`val`)) {
map[`val`]!!
} else {
`val`
}
}
}
/*
* Your Solution object will be instantiated and called as such:
* var obj = Solution(n, blacklist)
* var param_1 = obj.pick()
*/