528. Random Pick with Weight

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LeetCode

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Problem

You are given a 0-indexed array of positive integers w where w[i] describes the weight of the i^th index.

You need to implement the function pickIndex(), which randomly picks an index in the range [0, w.length - 1] (inclusive) and returns it. The probability of picking an index i is w[i] / sum(w).

• For example, if w = [1, 3], the probability of picking index 0 is 1 / (1 + 3) = 0.25 (i.e., 25%), and the probability of picking index 1 is 3 / (1 + 3) = 0.75 (i.e., 75%).

Example 1:

Input
["Solution","pickIndex"]
[[[1]],[]]
Output
[null,0]

Explanation
Solution solution = new Solution([1]);
solution.pickIndex(); // return 0. The only option is to return 0 since there is only one element in w.

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 is returning the second element (index = 1) that has a probability of 3/4.
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 0. It is returning the first element (index = 0) that has a probability of 1/4.

Since this is a randomization problem, multiple answers are allowed.
All of 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 <= 10^4
• 1 <= w[i] <= 10^5
• pickIndex will be called at most 10^4 times.

Code

class Solution {
    int[] sum;

    public Solution(int[] w) {
        sum = new int[w.length];

        sum[0] = w[0];
        for(int i = 1; i < w.length; i++) {
            sum[i] = sum[i - 1] + w[i];
        }
    }

    public int pickIndex() {
        double target = Math.random() * sum[sum.length - 1];

        int left = 0;
        int right = sum.length - 1;

        while(left + 1 < right) {
            int mid = (left + right) / 2;

            if(sum[mid] == target) return mid;

            if(sum[mid] > target) {
                right = mid;
            } else {
                left = mid;
            }
        }

        if(sum[left] >= target) {
            return left;
        } else {
            return right;
        }
    }
}