README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2100-2199/2154.Keep%20Multiplying%20Found%20Values%20by%20Two/README_EN.md	1235	Weekly Contest 278 Q1	Array Hash Table Sorting Simulation

2154. Keep Multiplying Found Values by Two

中文文档

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

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 $\textit{nums}$.

Python3

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

Java

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;
    }
}

C++

class Solution {
public:
    int findFinalValue(vector<int>& nums, int original) {
        unordered_set<int> s(nums.begin(), nums.end());
        while (s.contains(original)) {
            original <<= 1;
        }
        return original;
    }
};

Go

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

TypeScript

function findFinalValue(nums: number[], original: number): number {
    const s: Set<number> = new Set([...nums]);
    while (s.has(original)) {
        original <<= 1;
    }
    return original;
}