528. Random Pick with Weight

按权重随机选择

You are given a 0-indexed array of positive integers w where w[i] describes the weight of the i^th 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 <= 10⁴
1 <= w[i] <= 10⁵
pickIndex will be called at most 10⁴ times.

分析：

通过前缀和数组 + 二分查找实现按权重随机选择，确保每个下标被选中的概率与其权重成正比。前缀和数组将权重映射为连续区间，每个下标 i 对应区间 [prefix_sum[i-1] + 1, prefix_sum[i]]。

import random
from itertools import accumulate


class Solution:

    def __init__(self, w: List[int]):
        self.presum = list(accumulate(w))
        self.w = w

    def pickIndex(self) -> int:
        i = random.randint(1, self.presum[-1])
        return bisect.bisect_left(self.presum, i)
        

# Your Solution object will be instantiated and called as such:
# obj = Solution(w)
# param_1 = obj.pickIndex()

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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.

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.

import random from itertools import accumulate class Solution: def __init__(self, w: List[int]): self.presum = list(accumulate(w)) self.w = w def pickIndex(self) -> int: i = random.randint(1, self.presum[-1]) return bisect.bisect_left(self.presum, i) # Your Solution object will be instantiated and called as such: # obj = Solution(w) # param_1 = obj.pickIndex()