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.
classSolution {Random random;int[] wSums;publicSolution(int[] w) {this.random=newRandom();for (int i =1; i <w.length; ++i) w[i] += w[i -1];this.wSums= w; }publicintpickIndex() {int len =wSums.length;int randomSum =random.nextInt(wSums[len -1]) +1;int left =0, right = len -1;// search positionwhile (left < right) {int mid = left + (right - left) /2;if (wSums[mid] == randomSum)return mid;elseif (wSums[mid] < randomSum) left = mid +1;else right = mid; }return left; }}