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

@lekzeey

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

利用均匀分布的随机数：

rand.nextInt(count) < K

代表了 k / count的概率新的index i会替换已有的sample。

(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