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[0] is equal to 1.
solution.pick(1);

Three types of answer:

1. Map Solution, O(N) memory, O(N) init, O(1) pick.

2. Like @dettier's Reservoir Sampling. O(1) init, O(1) memory, but O(N) to pick.

Reservior Sampling

public class Solution {
    int[] arr;
    Random random;
    

    public Solution(int[] nums) {
        arr = nums;
        random = new Random();
    }
    
    public int pick(int target) {
        int count = 0;
        int index = 0;
        for (int i=0; i<arr.length; i++) {
            if (arr[i] == target) {
                count++;
                if (random.nextInt(count) == 0) {
                    index = i;
                }
            }
        }
        return index;
    }
}

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(nums);
 * int param_1 = obj.pick(target);
 */

 

Map solution: MLE for big case

public class Solution {

    public Solution(int[] nums) {
        for (int i=0; i<nums.length; i++) {
            int num = nums[i];
            if (!indexes.containsKey(num))
                indexes.put(num, new ArrayList<Integer>());
            indexes.get(num).add(i);
        }
    }
    
    public int pick(int target) {
        List<Integer> indexes = this.indexes.get(target);
        int i = (int) (Math.random() * indexes.size());
        return indexes.get(i);
    }
    
    private Map<Integer, List<Integer>> indexes = new HashMap<>();
}