chore: add LeetCode daily solution

Author: TechnoBlogger14o3

Easy/2025-11-19-2154-Keep-Multiplying-Found-Values-by-Two/explanation.md

@@ -0,0 +1,28 @@
+# Keep Multiplying Found Values by Two (Easy)
+
+**Problem ID:** 2154  
+**Date:** 2025-11-19  
+**Link:** https://leetcode.com/problems/keep-multiplying-found-values-by-two/
+
+## Approach
+
+To solve the problem "Keep Multiplying Found Values by Two," we will adopt a straightforward iterative approach that utilizes a set for efficient lookups.
+
+### Main Idea:
+The core idea is to repeatedly check if the current value of `original` exists in the given array `nums`. If it does, we multiply `original` by two and continue checking. If it does not exist, we stop the process and return the current value of `original`.
+
+### Steps:
+1. **Convert `nums` to a Set**: Since we need to perform multiple membership checks, converting the list `nums` to a set will allow for O(1) average time complexity for lookups.
+  
+2. **Iterate Until Not Found**: Initialize a loop that continues as long as `original` is found in the set of `nums`. Inside the loop, if `original` is found, multiply it by two.
+
+3. **Return Result**: Once `original` is no longer found in `nums`, exit the loop and return the final value of `original`.
+
+### Data Structures:
+- **Set**: We will use a set to store the elements of `nums` for efficient membership testing.
+
+### Complexity:
+- **Time Complexity**: O(n) for converting the list to a set, where n is the length of `nums`. The while loop will run at most log(1000) times (since `original` can double until it exceeds 1000), resulting in an overall time complexity of O(n).
+- **Space Complexity**: O(n) for storing the elements of `nums` in a set.
+
+This approach is efficient given the constraints and ensures that we can quickly determine whether to continue multiplying `original`.

Easy/2025-11-19-2154-Keep-Multiplying-Found-Values-by-Two/solution.java

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+class Solution {
+    public int findFinalValue(int[] nums, int original) {
+        Set<Integer> numSet = new HashSet<>();
+        for (int num : nums) {
+            numSet.add(num);
+        }
+        
+        while (numSet.contains(original)) {
+            original *= 2;
+        }
+        
+        return original;
+    }
+}

Easy/2025-11-19-2154-Keep-Multiplying-Found-Values-by-Two/solution.js

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+var findFinalValue = function(nums, original) {
+    const numSet = new Set(nums);
+    while (numSet.has(original)) {
+        original *= 2;
+    }
+    return original;
+};

Easy/2025-11-19-2154-Keep-Multiplying-Found-Values-by-Two/solution.py

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+class Solution:
+    def findFinalValue(self, nums: List[int], original: int) -> int:
+        num_set = set(nums)
+        while original in num_set:
+            original *= 2
+        return original

README.md

@@ -283,3 +283,4 @@ Through completing the Blind 75 and NeetCode 150, you will have mastered:
 - 2025-11-16 — [Number of Substrings With Only 1s](https://leetcode.com/problems/number-of-substrings-with-only-1s/) (Medium) → `Medium/2025-11-16-1513-Number-of-Substrings-With-Only-1s`
 - 2025-11-17 — [Check If All 1's Are at Least Length K Places Away](https://leetcode.com/problems/check-if-all-1s-are-at-least-length-k-places-away/) (Easy) → `Easy/2025-11-17-1437-Check-If-All-1-s-Are-at-Least-Length-K-Places-Away`
 - 2025-11-18 — [1-bit and 2-bit Characters](https://leetcode.com/problems/1-bit-and-2-bit-characters/) (Easy) → `Easy/2025-11-18-717-1-bit-and-2-bit-Characters`
+- 2025-11-19 — [Keep Multiplying Found Values by Two](https://leetcode.com/problems/keep-multiplying-found-values-by-two/) (Easy) → `Easy/2025-11-19-2154-Keep-Multiplying-Found-Values-by-Two`