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);

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 = 0; i < nums.length; i++) {
            if (target == nums[i]) {
                idxs.add(i);
            }
        }
        return idxs.get(r.nextInt(idxs.size()));
    }
}

For those who still didn't understand this, you could first do a count, and then generate random number.

public int pick(int target) {
    int count = 0;
    for(int i = 0; i < a.length; i++) {
        if(target == a[i]) count++;
    }
    int pickIndex = rand.nextInt(count);
    for(int i = 0; i < a.length; i++) {
        if(target == a[i]) {
            if(pickIndex == 0) return i;
               pickIndex--;
        }
    }
    return 0; // shouldn't come here
}

 

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 = -1;
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != target)
                continue;
            if (rnd.nextInt(++count) == 0)
                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.