Random Pick Index

Problem

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);

Solution

Using Random and nextInt(count) Solution

public class Solution {

    int[] nums;
    Random random;

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

    public int pick(int target) {
        int res = 0, count = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != target) continue;
            if (random.nextInt(++count) == 0) res = i;
        }
        return res;
    }
}

Analysis

To make each index has same probability to get called, we use Random.next(end)
Each time we meet the targeted number, we check random.nextInt(++count) == 0
Say there are two targeted numbers
First time we nextInt(1) will return 0 for sure since 1 is inclusive, and result = i
In the second time, nextInt(2) makes sure that current i and res each has 50% percent to be picked