# 398. Reservoir Sampling

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);
```

```cpp
class Solution {
    vector<int> nums;
public:
    Solution(vector<int> nums) {
        this->nums = nums;
    }

    int pick(int target) {
        int res;
        int cnt = 0;
        for(int i = 0; i < nums.size(); i++){
            if(nums[i] != target) continue;
            if(cnt == 0) {res = i; cnt++;}
            else{
                // with prob 1/(n+1) to replace the previous index
                if(rand()%(++cnt) == 0) res = i;
            }
        }

        return res;
    }
};

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

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

    public int pick(int target) {
        int cnt = 0, res = -1;
        for(int i = 0; i < nums.length; i++){
            if(nums[i] != target) continue;
            // with prob 1/(n+1) to replace the previous index
            if(rand.nextInt(++cnt) == 0)  res = i;
        }

        return res;
    }
}

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