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Random Pick with Weight

描述​

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.

分析​

先构造一个累加的概率数组，然后用二分查找。

代码​

// Random Pick with Weight
class Solution {
    private double[] p; // probability of each element

    // Time Complexity: O(n), Space Complexity: O(n)
    public Solution(int[] w) {
        double sum = 0;
        for(int x : w) sum += x;

        this.p = new double[w.length];
        for(int i = 0; i < w.length; i++){
            w[i] += (i == 0) ? 0 : w[i - 1];
            p[i] = w[i] / sum;
        }
    }

    // Time Complexity: O(logn), Space Complexity: O(1)
    public int pickIndex() {
        // upper bound
        return Math.abs(Arrays.binarySearch(this.p, Math.random())) - 1;
    }
}

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