Keep Multiplying Found Values by Two

Problem 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 in nums, multiply it by two (i.e., set original = 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

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[i], original <= 1000

Solution Approach

Intuition:

To efficiently Keep Multiplying Found Values by Two

Solution Implementation

Code In Different Languages:

JavaScript

Written by @Ishitamukherjee2004

 
class Solution {
findFinalValue(nums, original) {
 let n = nums.length;
 let i = 0;
 while(i < n){
   if(nums[i] === original){
     original *= 2;
     i = 0;
   } else {
     i++;
   }
 }
 return original;
}
}

TypeScript

Written by @Ishitamukherjee2004

 
class Solution {
findFinalValue(nums: number[], original: number): number {
 let n = nums.length;
 let i = 0;
 while(i < n){
   if(nums[i] === original){
     original *= 2;
     i = 0;
   } else {
     i++;
   }
 }
 return original;
}
}

Python

Written by @Ishitamukherjee2004

 class Solution:
 def findFinalValue(self, nums, original):
     n = len(nums)
     i = 0
     while i < n:
         if nums[i] == original:
             original *= 2
             i = 0
         else:
             i += 1
     return original

Java

Written by @Ishitamukherjee2004

 public class Solution {
 public int findFinalValue(int[] nums, int original) {
     int n = nums.length;
     int i = 0;
     while(i < n){
         if(nums[i] == original){
             original *= 2;
             i = 0;
         } else {
             i++;
         }
     }
     return original;
 }
}

C++

Written by @Ishitamukherjee2004

class Solution {
public:
 int findFinalValue(vector<int>& nums, int original) {
     int n = nums.size();
     int i = 0;
     while(i<n){
     if(nums[i]==original){
         original*=2;
         i=0;
     }
     else i++;

     }
     return original;
 }
};

Complexity Analysis

• Time Complexity: $O(m*n)$
• Space Complexity: $O(1)$
• The time complexity is $O(log(n))$ where n is the length of the input array nums. This is because in the worst case, the algorithm may need to iterate through the entire array for each doubling of original, resulting in a linear time complexity.
• The space complexity is $O(1)$ because we are not using any extra space.