# Random Pick Index

`Reservoir Sampling`

Given an array of integers with possible duplicates, randomly output the index of a given target number. You can assume that the given target number must exist in the array.

Note:
The array size can be very large. Solution that uses too much extra space will not pass the judge.

Example:

``````int[] nums = new int[] {1,2,3,3,3};
Solution solution = new Solution(nums);

// pick(3) should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.
solution.pick(3);

// pick(1) should return 0. Since in the array only nums[0] is equal to 1.
solution.pick(1);
``````

## Analysis

To those who don't understand why it works. Consider the example in the OJ
{1,2,3,3,3} with target 3,you want to select 2,3,4 with a probability of 1/3 each.

2 : It's probability of selection is 1 * (1/2) * (2/3) = 1/3
3 : It's probability of selection is (1/2) * (2/3) = 1/3
4 : It's probability of selection is just 1/3

``````rand.nextInt(count) < K
``````

(1 / i) * (1 - 1/ (i + 1)) * (1 - 1/(i + 2)) * ... * (1 - 1 / n) = 1/n

## Solution

#### Reservoir Sampling

121 ms, faster than 94.95%

``````class Solution {
private Random rand;
private int[] nums;
private int K = 1;

public Solution(int[] nums) {
rand = new Random();
this.nums = nums;
}

public int pick(int target) {
int result = -1;
int count = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] == target) {
count++;
if (rand.nextInt(count) < K) {
result = i;
}
}
}
return result;
}
}

/**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(nums);
* int param_1 = obj.pick(target);
*/
``````

## Reference

https://leetcode.com/problems/random-pick-index/discuss/88072/Simple-Reservoir-Sampling-solution