leetcode_528
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).
Example 1:
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:
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.
Solutions
prefix weight sum
class Solution {
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);
}
int pickIndex() {
int rn = uni(gen);
return upper_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();
*/
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