[LeetCode] 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);

Thought process:
The question doesn't have a required time complexity on pick, but rather it doesn't allow using too much extra space. The idea is to store the given array as is. Iterate through the array. Use a variable count to keep track of the number of occurrences of target. And use count as an upper bound for random.

Solution:

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class Solution {
    private int[] reservoir;
    private Random random;

    public Solution(int[] nums) {
        reservoir = nums;
        random = new Random();
    }
    
    public int pick(int target) {
        int result = -1;
        int count = 0;
        
        for (int i = 0; i < reservoir.length; i++) {
            if (reservoir[i] != target) {
                continue;
            }
            count++;
            if (random.nextInt(count) == 0) {
                result = i;
            }
        }
        return result;
    }
}

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

Time complexity:
pick(): O(n).
Location: Philadelphia, PA, USA

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