Welcome to Subscribe On Youtube
2154. Keep Multiplying Found Values by Two
Description
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 innums
, multiply it by two (i.e., setoriginal = 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
.
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.
Constraints:
1 <= nums.length <= 1000
1 <= nums[i], original <= 1000
Solutions

class Solution { public int findFinalValue(int[] nums, int original) { Set<Integer> s = new HashSet<>(); for (int num : nums) { s.add(num); } while (s.contains(original)) { original <<= 1; } return original; } }

class Solution { public: int findFinalValue(vector<int>& nums, int original) { unordered_set<int> s; for (int num : nums) s.insert(num); while (s.count(original)) original <<= 1; return original; } };

class Solution: def findFinalValue(self, nums: List[int], original: int) > int: s = set(nums) while original in s: original <<= 1 return original

func findFinalValue(nums []int, original int) int { s := make(map[int]bool) for _, num := range nums { s[num] = true } for s[original] { original <<= 1 } return original }

function findFinalValue(nums: number[], original: number): number { let set: Set<number> = new Set(nums); while (set.has(original)) { original *= 2; } return original; }