leetcode - 528. Random Pick with Weight
Description
You are given a 0-indexed array of positive integers w where w[i] describes the weight of the ith index.
You need to implement the function pickIndex(), which randomly picks an index in the range [0, w.length - 1] (inclusive) and returns it. The probability of picking an index i is w[i] / sum(w).
For example, if w = [1, 3], the probability of picking index 0 is 1 / (1 + 3) = 0.25 (i.e., 25%), and the probability of picking index 1 is 3 / (1 + 3) = 0.75 (i.e., 75%).
Example 1:
Input
["Solution","pickIndex"]
[[[1]],[]]
Output
[null,0]
Explanation
Solution solution = new Solution([1]);
solution.pickIndex(); // return 0. The only option is to return 0 since there is only one element in w.
Example 2:
Input
["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"]
[[[1,3]],[],[],[],[],[]]
Output
[null,1,1,1,1,0]
Explanation
Solution solution = new Solution([1, 3]);
solution.pickIndex(); // return 1. It is returning the second element (index = 1) that has a probability of 3/4.
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 0. It is returning the first element (index = 0) that has a probability of 1/4.
Since this is a randomization problem, multiple answers are allowed.
All of the following outputs can be considered correct:
[null,1,1,1,1,0]
[null,1,1,1,1,1]
[null,1,1,1,0,0]
[null,1,1,1,0,1]
[null,1,0,1,0,0]
......
and so on.
Constraints:
1 <= w.length <= 10^4
1 <= w[i] <= 10^5
pickIndex will be called at most 104 times.
Solution
Solved after help.
Use accumulated frequency, for example, when weights = [2,5,3,4]
, accumulated frequency is [2,7,10,14]
, and generate a random number from [1, 14]
inclusively. Then:
index in [1, 2]
, the result is 0
index in [3, 7]
, the result is 1
index in [8, 10]
, the result is 2
index in [11, 14]
, the result is 3
Use binary search to find where the index falls.
Time complexity:
o
(
n
)
o(n)
o(n) for init,
o
(
log
?
n
)
o(\log n)
o(logn) for pick
Space complexity:
o
(
n
)
o(n)
o(n)
Code
class Solution:
def __init__(self, w: List[int]):
self.weights = []
for wi in w:
if not self.weights:
self.weights.append(wi)
else:
new_weight = wi + self.weights[-1]
self.weights.append(new_weight)
def pickIndex(self) -> int:
random_i = random.choice(range(1, self.weights[-1] + 1))
left, right = 0, len(self.weights) - 1
while left < right:
mid = (left + right) >> 1
if random_i <= self.weights[mid]:
right = mid
else:
left = mid + 1
return (left + right) >> 1
# Your Solution object will be instantiated and called as such:
# obj = Solution(w)
# param_1 = obj.pickIndex()
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