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);

 public class Solution {

     int[] nums;
     Random r = new Random();

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

     public int pick(int target) {
         ArrayList<Integer> idxs = new ArrayList<Integer>();
         for (int i = ; i < nums.length; i++) {
             if (target == nums[i]) {
                 idxs.add(i);
             }
         }
         return idxs.get(r.nextInt(idxs.size()));
     }
 }

Simple Reservior Sampling approach

 public class Solution {

     int[] nums;
     Random rnd;

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

     public int pick(int target) {
         int result = -;
         int count = ;
         for (int i = ; i < nums.length; i++) {
             if (nums[i] != target)
                 continue;
             if (rnd.nextInt(++count) == )
                 result = i;
         }

         return result;
     }
 }

Simple Reservior Sampling

Suppose we see a sequence of items, one at a time. We want to keep a single item in memory, and we want it to be selected at random from the sequence. If we know the total number of items (n), then the solution is easy: select an index ibetween 1 and n with equal probability, and keep the i-th element. The problem is that we do not always know n in advance. A possible solution is the following:

• Keep the first item in memory.
• When the i-th item arrives (for {\displaystyle i>1}):
  • with probability {\displaystyle 1/i}, keep the new item (discard the old one)
  • with probability {\displaystyle 1-1/i}, keep the old item (ignore the new one)

So:

• when there is only one item, it is kept with probability 1;
• when there are 2 items, each of them is kept with probability 1/2;
• when there are 3 items, the third item is kept with probability 1/3, and each of the previous 2 items is also kept with probability (1/2)(1-1/3) = (1/2)(2/3) = 1/3;
• by induction, it is easy to prove that when there are n items, each item is kept with probability 1/n.