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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:

  1. If original is found in nums, multiply it by two (i.e., set original = 2 * original).
  2. Otherwise, stop the process.
  3. 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;
}

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