You are given an array of integers nums. You are also given an integer original which is the first number that needs to be searched for in nums.

You then do the following steps:


	If original is found in nums, multiply it by two (i.e., set original = 2 * original).
	Otherwise, stop the process.
	Repeat this process with the new number as long as you keep finding the number.


Return the final value of original.

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Example 1:

Input: nums = [5,3,6,1,12], original = 3
Output: 24
Explanation: 
- 3 is found in nums. 3 is multiplied by 2 to obtain 6.
- 6 is found in nums. 6 is multiplied by 2 to obtain 12.
- 12 is found in nums. 12 is multiplied by 2 to obtain 24.
- 24 is not found in nums. Thus, 24 is returned.


Example 2:

Input: nums = [2,7,9], original = 4
Output: 4
Explanation:
- 4 is not found in nums. Thus, 4 is returned.


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Constraints:


	1 &lt;= nums.length &lt;= 1000
	1 &lt;= nums[i], original &lt;= 1000


class Solution:
    def findFinalValue(self, nums: List[int], original: int) -> int:
        while original in nums:
            original *= 2
        return original
class Solution
{
    public int findFinalValue(int[] nums, int original)
    {
        HashSet<Integer> set = new HashSet<>();
        for(int i : nums)
            if(i >= original)
                set.add(i);
        while(true)
            if(set.contains(original))
                original *= 2;
            else
                break;
        return original;
    }
}
class Solution {
public:
    int findFinalValue(vector<int>& nums, int original) {
        int n = 1;
        for(int i = 0; i<n;++i)
        {
            if(find(nums.begin(),nums.end(),original) != nums.end())    //find func detailled explanation above
            {
                original *= 2;
                n += 1; //n is incremented by one beacuse questions want us to perform the operation again if element is found again after its double.
            }
        }
        return original;
    }
};
var findFinalValue = function(nums, original) {
    while (nums.includes(original)) {
        original = original * 2
    }

    return original
};
Keep Multiplying Found Values by Two