diff --git a/Java/0983.java b/Java/0983.java
index fa451bc..69b30eb 100644
--- a/Java/0983.java
+++ b/Java/0983.java
@@ -1,31 +1,15 @@
-class Solution {
-    int[] days, costs;
-    Integer[] memo;
-    int[] durations = new int[]{1, 7, 30};
-
-    public int mincostTickets(int[] days, int[] costs) {
-        this.days = days;
-        this.costs = costs;
-        memo = new Integer[days.length];
-
-        return dp(0);
-    }
-
-    public int dp(int i) {
-        if (i >= days.length)
-            return 0;
-        if (memo[i] != null)
-            return memo[i];
-
-        int ans = Integer.MAX_VALUE;
-        int j = i;
-        for (int k = 0; k < 3; ++k) {
-            while (j < days.length && days[j] < days[i] + durations[k])
-                j++;
-            ans = Math.min(ans, dp(j) + costs[k]);
-        }
-
-        memo[i] = ans;
-        return ans;
-    }
-}
+class Solution {
+    public int mincostTickets(int[] days, int[] costs) {
+        int n = days.length;
+        int[] dp = new int[n+1];
+        for(int i = 0; i < n; ++i) dp[i] = 365*costs[0];
+        
+        for(int i = n-1; i >= 0; --i){
+            int d7 = i, d30 = i;
+            while(d7 < n && days[d7] < days[i] + 7) ++d7;
+            while(d30 < n && days[d30] < days[i] + 30) ++d30;
+            dp[i] = Math.min(costs[0] + dp[i+1], Math.min(costs[1] + dp[d7], costs[2] + dp[d30]));
+        }
+        return dp[0];
+    }
+}
diff --git a/README.md b/README.md
index 864760b..255d708 100644
--- a/README.md
+++ b/README.md
@@ -38,6 +38,7 @@ Or from the Sheet 1 of this [Google Sheet](https://bit.ly/2EUhwnw)
 | 0695 | [Max Area of Island](https://leetcode.com/problems/max-area-of-island/) | Apply DFS on the Gird by exploring every square connected to it 4-directionally, total number of squares explored will be the area of that connected shape. | [C++](https://github.com/sankalpdayal5/LeetCode-Solutions/blob/master/C++/0695.cpp)     | O(R∗C) | O(R∗C)  | Medium     | Depth-First-Search         |    |
 | 0876 | [Middle of the Linked List](https://leetcode.com/problems/middle-of-the-linked-list/solution/) | Find the length of given linked list and then traverse from root till one less than half of linked list length and print the next node to current node | [Java](https://github.com/sankalpdayal5/LeetCode-Solutions/blob/master/Java/0876.java)     | O(n) | O(1)  | Easy     | Linked List        |    |
 | 0983 | [Minimum Cost For Tickets](https://leetcode.com/problems/minimum-cost-for-tickets/)                    | Here dp(i) is the cost to travel from day days[i] to the end of the plan. if days[j] < days[i] + 1 then j1=j. if days[j] < days[i] + 7 then j=j7. if days[j] < days[i] + 30 then j=j30 . dp(i)=min(dp(j1)+costs[0],dp(j7)+costs[1],dp(j30)+costs[2])    | [Java](https://github.com/sankalpdayal5/LeetCode-Solutions/blob/master/Java/0983.java)            | O(n) | O(n)  | Medium       | Dynamic Programming |    [📺](https://www.youtube.com/watch?v=2AnrAlCA578)    |
+| 0983 | [Minimum Cost For Tickets]https://leetcode.com/problems/minimum-cost-for-tickets/solution/) |    | [Java](https://github.com/sankalpdayal5/LeetCode-Solutions/blob/master/Java/0983.java)            | O(n) | O(n)  | Medium       | Dynamic Programming |    [📺](https://www.youtube.com/watch?v=2AnrAlCA578)    |
 | 1347 | [Minimum Number of Steps to Make Two Strings Anagram](https://leetcode.com/problems/minimum-number-of-steps-to-make-two-strings-anagram/) | Add 1 for char in s and remove 1 for char in t | [Java](https://github.com/sankalpdayal5/LeetCode-Solutions/blob/master/Java/1347.java)     [Python](https://github.com/sankalpdayal5/LeetCode-Solutions/blob/master/Python/1347.py) | O(n+m) | O(1)  | Medium     | Hash Table Heap         | [📺](https://www.youtube.com/watch?v=xXXOpOYWtRE)    |
 | 1352 | [Product of the Last K Numbers](https://leetcode.com/problems/product-of-the-last-k-numbers/) | Add new element to list by multiplying it with previous number and return arr[n-1]/arr[n-k-1] | [Python](https://github.com/sankalpdayal5/LeetCode-Solutions/blob/master/Python/1352.py)      | O(1) | O(1)  | Medium     | Array Design         | [📺](https://www.youtube.com/watch?v=8CuVduv0Kyg)    |