# 528. Random Pick with Weight

## 1. Question

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).

## 2. Examples

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.


## 3. Constraints

• 1 <= w.length <= 10000
• 1 <= w[i] <= 10^5
• pickIndex will be called at most 10000 times.

## 5. Solutions

class Solution {

int[] arr;
int sum;

public Solution(int[] w) {
arr = new int[w.length];
arr[0] = w[0];
for (int i = 1; i < w.length; i++) {
arr[i] = arr[i - 1] + w[i];
}
sum = Arrays.stream(arr).sum();
}

public int pickIndex() {
return binarySearch((int) (Math.random() * arr[arr.length - 1] + 1));
}

public int binarySearch(int num) {
int left = 0;
int right = arr.length - 1;
while (left < right) {
int mid = (right - left) / 2 + left;
if (arr[mid] < num) {
left = mid + 1;
} else {
right = mid;
}
}
return left;
}

}

/**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(w);
* int param_1 = obj.pickIndex();
*/
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