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
Solution 1: Hash Table
We use a hash table $\textit{s}$ to record all the numbers in the array $\textit{nums}$.
Next, starting from $\textit{original}$, if $\textit{original}$ is in $\textit{s}$, we multiply $\textit{original}$ by $2$ until $\textit{original}$ is not in $\textit{s}$ anymore, then return $\textit{original}$.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.
1 2 3 4 5 6 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
|
1 2 3 4 5 6 7 8 9 10 |
|
1 2 3 4 5 6 7 8 9 10 |
|
1 2 3 4 5 6 7 |
|