0528. Random Pick with Weight

https://leetcode.com/problems/random-pick-with-weight

Description

You are given an array of positive integers w where w[i] describes the weight of i``thindex (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.

ac: prefix sum

class Solution {
    int[] prefixSum;
    int upperLimit;
    Random random;
    
    public Solution(int[] w) {
        prefixSum = new int[w.length + 1];
        for (int i = 1; i < prefixSum.length; i++) {
            prefixSum[i] = prefixSum[i-1] + w[i-1];
        }
        upperLimit = prefixSum[prefixSum.length-1];
        int[] test = prefixSum;
        random = new Random();
    }
    
    public int pickIndex() {
        int r = random.nextInt(upperLimit);
        int index = Arrays.binarySearch(prefixSum, r); // Return insertion index if exists, otherwise -(insertion index)-1;
        return index >= 0 ? index : -index-1-1; // in "-(insertion index)-1" insertion index is the index of it after the insertion, so we find previous index.
    }
}