398. Random Pick Index

398. Random Pick Index

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);

My Solutions:

Iterate through the element in array and once find, increment the count, and determine whether random nextInt(count) == 0

(Resorvior sampling: the probability to choose an element out of n elements is 1/n)

count代表总共有几个target出现，每个target出现的几率应该是1/count

class Solution {

    private int[] nums;
    
    public Solution(int[] nums) {
        this.nums = nums;
    }
    
    public int pick(int target) {
        int index = -1, count = 0;
        Random r = new Random();
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == target) {
                count++;
                if (r.nextInt(count) == 0) index = i;
            }
        }
        return index;
    }
}