Random Pick With Weight Problem


Description

LeetCode Problem 528.

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:

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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.

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 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 10^4 times.


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class Solution {
public:
    vector<int> s;
    Solution(vector<int> w) {
        for (int i : w) s.push_back(s.empty() ? i : (i + s.back()));
    }
    
    int pickIndex() {
        int m = s.back();
        int r = rand() % m;
        auto it = upper_bound(s.begin(), s.end(), r);
        return it - s.begin();
    }
};




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