Random Pick with Weight
Given an array of positive integers w. where w[i]
describes the weight of i
th
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 index 0
is 1 / (1 + 3) = 0.25
(i.e 25%) while the probability of picking 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:
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.
class Solution {
private:
vector<int> v;
int s;
public:
Solution(vector<int>& w) {
v.push_back(w[0]);
for(int i=1;i<w.size();++i)
v.push_back(w[i]+v.back());
s=w.size();
}
int pickIndex() {
int i=(rand()%v[s-1])+1;
int l=0,r=s-1;
while(l<r)
{
int mid=(r+l)/2;
if(v[mid]==i)
return mid;
else if(v[mid]<i)
l=mid+1;
else
r=mid;
}
return l;
}
};
/**
* Your Solution object will be instantiated and called as such:
* Solution* obj = new Solution(w);
* int param_1 = obj->pickIndex();
*/