LeetCode Challenge Day 65 — 2154. Keep Multiplying Found Values by Two

Hey folks 👋

This is Day 65 of my LeetCode streak 🚀
Today’s problem is 2154 — Keep Multiplying Found Values by Two.

A fun simulation-style problem:
You repeatedly check whether your current number exists in the array —
if yes, double it and continue the process.

💡 Intuition

The straightforward idea is:

• If the number is present → multiply by 2
• Otherwise → stop

However, repeatedly searching through the array is slow.
So to optimize, we use a Set, allowing constant-time lookups.

📌 Approach

1. Convert the input nums into a Set for O(1) search.
2. While the current original value exists inside the Set:
   • Update original = original * 2.
3. When the number is no longer present, return the final value.

This ensures minimal overhead and avoids unnecessary repeated scans.

📈 Complexity

• Time Complexity: O(n)
  Creating the Set takes O(n); the doubling process is very short.

• Space Complexity: O(n)
  A Set is used to store the list values.

🧑‍💻 Code (JavaScript)

var findFinalValue = function(nums, original) {
    const set = new Set(nums);

    while (set.has(original)) {
        original *= 2;
    }

    return original;
};

🎯 Example

Input:
nums = [5,3,6,1,12], original = 3
Output:
24

Explanation:
3 is in nums → becomes 6
6 is in nums → becomes 12
12 is in nums → becomes 24
24 is not in nums, so we stop.

🧠 Reflection

A clean demonstration of how hashing simplifies repeated search tasks.
By converting to a Set, the problem becomes a simple loop instead of repeated array scans.

See you tomorrow for Day 66! 🚀
Happy Coding 👨‍💻✨