diff --git a/Leetcode/Min_Cost_Climbing_Stairs.cpp b/Leetcode/Min_Cost_Climbing_Stairs.cpp
index c7ffc5b..84a85c6 100644
--- a/Leetcode/Min_Cost_Climbing_Stairs.cpp
+++ b/Leetcode/Min_Cost_Climbing_Stairs.cpp
@@ -1,13 +1,41 @@
 class Solution {
 public:
-    int minCostClimbingStairs(vector<int>& cost) {
-        int n=cost.size();
-        int dp[n];
-        dp[0]=cost[0];
-        dp[1]=cost[1];
-        for(int i=2;i<n;i++){
-            dp[i]=min(dp[i-1],dp[i-2])+cost[i];
+    int dp[1000];
+    int rec(int level, int size, vector<int> &cost)
+    {
+//         this recurrence gives us the minimum cost of reaching the top of the floor
+            
+//         pruning
+        if(level > size)
+        {
+            return 1001;
+//             we will return anything > 999 so that this partial-solution is never chosen
+        }
+//         base case
+        
+        if(level == size)
+        {
+            return 0;
+        }
+        
+        if(dp[level] != -1)
+        {
+            return dp[level];
         }
-        return min(dp[n-1],dp[n-2]);
+        
+//      choices
+//         choices will be to either take 1 step or two steps
+        
+        int ans = 0;
+        
+        ans = cost[level] + min(rec(level + 1, size, cost), rec(level + 2, size, cost));
+        
+//         return and save
+        dp[level] = ans;
+        return ans;
+    }
+    int minCostClimbingStairs(vector<int>& cost) {
+        memset(dp, -1, sizeof(dp));
+        return min(rec(1, cost.size(), cost), rec(0, cost.size(), cost));   
     }
-};
+};
\ No newline at end of file