 # [Leetcode] 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 is equal to 1. solution.pick(1);```
• Create a Map for every target -> range[]. Now, given a target we can query its range in O(1) time. Note, that length of this range is simply range-range+1.
• Given a range (for any target), generate a random_number between 0 to length_of_range (Java’s Random class allows us to generate a random number between 0 (inclusive) to n (exclusive)). We then simple return i + random_number.
```class Solution {   Map<Integer, int[]> rangeMap; Random r;   public Solution(int[] nums) { r = new Random(); rangeMap = new HashMap<Integer, int[]>(); int i = 0; while ( i < nums.length ) { int num = nums[i]; int j = i; while ( j < nums.length && nums[j] == num ) j++; int range[] = { i, j-1 }; rangeMap.put(num, range); i = j; } }   public int pick(int target) { int range[] = rangeMap.get(target); return range + r.nextInt(range-range+1); } }   /** * Your Solution object will be instantiated and called as such: * Solution obj = new Solution(nums); * int param_1 = obj.pick(target); */```