Formatted question description: https://leetcode.ca/all/2154.html

2154. Keep Multiplying Found Values by Two (Easy)

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

Similar Questions:

Solution 1. Set + Simulation

// OJ: https://leetcode.com/problems/keep-multiplying-found-values-by-two/

// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int findFinalValue(vector<int>& A, int n) {
        set<int> s(begin(A), end(A));
        while (s.count(n)) n *= 2;
        return n;
    }
};

Solution 2. Bucket Count + Simulation

// OJ: https://leetcode.com/problems/keep-multiplying-found-values-by-two/

// Time: O(N)
// Space: O(R) where `R` is the range of numbers in `A`
class Solution {
public:
    int findFinalValue(vector<int>& A, int n) {
        int seen[1001] = {};
        for (int x : A) seen[x] = 1;
        while (n < 1001 && seen[n]) n *= 2;
        return n;
    }
};

Solution 3. Sorting + One Pass

// OJ: https://leetcode.com/problems/keep-multiplying-found-values-by-two/

// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int findFinalValue(vector<int>& A, int n) {
        sort(begin(A), end(A));
        for (int i = 0, N = A.size(); i < N;) {
            while (i < N && A[i] < n) ++i;
            if (i == N || A[i] != n) break;
            n *= 2;
        }
        return n;
    }
};

Discuss

https://leetcode.com/problems/keep-multiplying-found-values-by-two/discuss/1730124

All Problems

All Solutions