leetcode - 528. Random Pick with Weight

发布时间:2023年12月20日

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()
文章来源:https://blog.csdn.net/sinat_41679123/article/details/135091208
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