diff --git a/Regular-Question-Answers/largest-divisible-subset.cpp b/Regular-Question-Answers/largest-divisible-subset.cpp
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+// Author: Racha Badreddine
+// Reviewer:
+// Question Link: https://leetcode.com/problems/largest-divisible-subset/
+
+// Time Complexity Analysis: O(n^2)
+// Sorting has O(n log n) complexity but is neglected 
+class Solution {
+public:
+    vector<int> largestDivisibleSubset(vector<int>& nums) {
+        // Sort the array 
+        sort(nums.begin(), nums.end());
+
+        // Variable to store the array size
+        int size = nums.size();
+
+        // Vector to hold the largest divisible subsets starting at each index i
+        vector<vector<int>> dp(size);
+
+        // Vector to hold the largest divisible subset
+        vector<int> result;
+
+        // Iterate through elements starting from the end
+        for(int i = size - 1; i >= 0; i--) {
+
+            // The smallest divisible subset contains the element itself
+            dp[i].push_back(nums[i]);
+
+            // Iterate through the elements larger than the current one
+            for(int j = i + 1; j < size; j++) {
+
+                // If the element is divisible AND if adding the current element to the subset already stored in the jth element is larger than our current subset 
+                if(nums[j] % nums[i] == 0 && (dp[j].size() + 1 > dp[i].size())) {
+                    // Copy the subset to the current element
+                    dp[i] = dp[j];
+                    // Insert the ith element to the beginning of the subset
+                    dp[i].insert(dp[i].begin(), nums[i]);
+                }
+            }
+
+            // If the current subset is larger than the previous one, copy it to the result
+            if(dp[i].size() > result.size()) result = dp[i];
+        }
+
+        // Return the largest divisible subset
+        return result;
+    }
+};
+