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:
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.
class Solution {
Random random;
int[] wSums;
public Solution(int[] w) {
this.random = new Random();
for(int i=1; i<w.length; ++i)
w[i] += w[i-1];
this.wSums = w;
}
public int pickIndex() {
int len = wSums.length;
int idx = random.nextInt(wSums[len-1]) + 1;
int left = 0, right = len - 1;
// search position
while(left < right){
int mid = left + (right-left)/2;
if(wSums[mid] == idx)
return mid;
else if(wSums[mid] < idx)
left = mid + 1;
else
right = mid;
}
return left;
}
}