# Leetcode Algorithms

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);`

Logic:

Create a random set, and then iterate through the set. If the value of the set is equal to target, increment count.

Solution:

`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;    }};`