Given an array w of positive integers, where w[i] describes the weight of index i(0-indexed), write a function pickIndex which randomly picks an index in proportion to its weight.
For example, given an input list of values w = [2, 8], when we pick up a number out of it, the chance is that 8 times out of 10 we should pick the number 1 as the answer since it's the second element of the array (w[1] = 8).
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:
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.
Solutions
prefix weight sum
classSolution{public: vector<int> psum; mt19937 gen{random_device{}()}; uniform_int_distribution<int> uni;Solution(vector<int>&w){int sum =0;for(auto n : w)psum.push_back(sum += n); uni =uniform_int_distribution<>(0, sum -1);}intpickIndex(){int rn =uni(gen);returnupper_bound(psum.begin(),psum.end(), rn)-psum.begin();}};/** * Your Solution object will be instantiated and called as such: * Solution* obj = new Solution(w); * int param_1 = obj->pickIndex();*/