398.md

398. Random Pick Index

Description

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);

Thinking Process

1. This is a typical reservoir sampling problem wiki link
2. Scan through the array, whenever we hit the target, increment the count by 1. The new index has a probability of 1/count to replace the old index.
3. Return the index

Code

public class Solution {
    int[] nums;
    Random rand;

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

    public int pick(int target) {
        int count = 0;
        int index = -1;
        for(int i = 0; i < nums.length; i++){
            if(nums[i] == target){
                index = (rand.nextInt(++count) == 0) ? i : index; 
            }
        }
        return index;
    }
}

Complexity

1. Space complexity is O(1) since no addtional data structures are used
2. Time complexity is O(n) since the array has to be scanned once for every pick() call