Problem description:

You are given an array of positive integers w where w[i] describes the weight of ith index (0-indexed).

We need to call the function pickIndex() which randomly returns an integer in the range [0, w.length - 1]. pickIndex() should return the integer proportional to its weight in the w array. For example, for w = [1, 3], the probability of picking the index 0 is 1 / (1 + 3) = 0.25 (i.e 25%) while the probability of picking the index 1 is 3 / (1 + 3) = 0.75 (i.e 75%).

More formally, the probability of picking index i is w[i] / sum(w).

Example 1:

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Input
["Solution","pickIndex"]
[[[1]],[]]
Output
[null,0]

Explanation
Solution solution = new Solution([1]);
solution.pickIndex(); // return 0. Since there is only one single element on the array the only option is to return the first element.

Example 2:

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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's returning the second element (index = 1) that has probability of 3/4.
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 0. It's returning the first element (index = 0) that has probability of 1/4.

Since this is a randomization problem, multiple answers are allowed so 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 <= 10000
  • 1 <= w[i] <= 10^5
  • pickIndex will be called at most 10000 times.

Solution:

The idea is to create a prefix_sum array and have a sum for the total weight.

Then multiply the sum with random when picking a number. Do a binary search to find which index has the number that is smaller but closest to the random*sum

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class Solution:

    def __init__(self, weight: List[int]):
        self.prefix_sum = []
        prefix = 0
        for w in weight:
            prefix += w
            self.prefix_sum.append(prefix)
        self.total_sum = prefix

    def pickIndex(self) -> int:
        rand = self.total_sum* random.random()
        i = bisect_left(self.prefix_sum, rand)
        return i

time complexity: $O(n)$ in init, $O(logn)$ when pickIndex
space complexity: $O(n)$
reference:
related problem: