Day 50 | 123. Best Time to Buy and Sell Stock III | 188. Best Time to Buy and Sell Stock IV

Day 1 | 704. Binary Search | 27. Remove Element | 35. Search Insert Position | 34. First and Last Position of Element in Sorted Array
Day 2 | 977. Squares of a Sorted Array | 209. Minimum Size Subarray Sum | 59. Spiral Matrix II
Day 3 | 203. Remove Linked List Elements | 707. Design Linked List | 206. Reverse Linked List
Day 4 | 24. Swap Nodes in Pairs| 19. Remove Nth Node From End of List| 160.Intersection of Two Lists
Day 6 | 242. Valid Anagram | 349. Intersection of Two Arrays | 202. Happy Numbe | 1. Two Sum
Day 7 | 454. 4Sum II | 383. Ransom Note | 15. 3Sum | 18. 4Sum
Day 8 | 344. Reverse String | 541. Reverse String II | 替换空格 | 151.Reverse Words in a String | 左旋转字符串
Day 9 | 28. Find the Index of the First Occurrence in a String | 459. Repeated Substring Pattern
Day 10 | 232. Implement Queue using Stacks | 225. Implement Stack using Queue
Day 11 | 20. Valid Parentheses | 1047. Remove All Adjacent Duplicates In String | 150. Evaluate RPN
Day 13 | 239. Sliding Window Maximum | 347. Top K Frequent Elements
Day 14 | 144.Binary Tree Preorder Traversal | 94.Binary Tree Inorder Traversal| 145.Binary Tree Postorder Traversal
Day 15 | 102. Binary Tree Level Order Traversal | 226. Invert Binary Tree | 101. Symmetric Tree
Day 16 | 104.MaximumDepth of BinaryTree| 111.MinimumDepth of BinaryTree| 222.CountComplete TreeNodes
Day 17 | 110. Balanced Binary Tree | 257. Binary Tree Paths | 404. Sum of Left Leaves
Day 18 | 513. Find Bottom Left Tree Value | 112. Path Sum | 105&106. Construct Binary Tree
Day 20 | 654. Maximum Binary Tree | 617. Merge Two Binary Trees | 700.Search in a Binary Search Tree
Day 21 | 530. Minimum Absolute Difference in BST | 501. Find Mode in Binary Search Tree | 236. Lowes
Day 22 | 235. Lowest Common Ancestor of a BST | 701. Insert into a BST | 450. Delete Node in a BST
Day 23 | 669. Trim a BST | 108. Convert Sorted Array to BST | 538. Convert BST to Greater Tree
Day 24 | 77. Combinations
Day 25 | 216. Combination Sum III | 17. Letter Combinations of a Phone Number
Day 27 | 39. Combination Sum | 40. Combination Sum II | 131. Palindrome Partitioning
Day 28 | 93. Restore IP Addresses | 78. Subsets | 90. Subsets II
Day 29 | 491. Non-decreasing Subsequences | 46. Permutations | 47. Permutations II
Day 30 | 332. Reconstruct Itinerary | 51. N-Queens | 37. Sudoku Solver
Day 31 | 455. Assign Cookies | 376. Wiggle Subsequence | 53. Maximum Subarray
Day 32 | 122. Best Time to Buy and Sell Stock II | 55. Jump Game | 45. Jump Game II
Day 34 | 1005. Maximize Sum Of Array After K Negations | 134. Gas Station | 135. Candy
Day 35 | 860. Lemonade Change | 406. Queue Reconstruction by Height | 452. Minimum Number of Arrows
Day 36 | 435. Non-overlapping Intervals | 763. Partition Labels | 56. Merge Intervals
Day 37 | 738. Monotone Increasing Digits | 714. Best Time to Buy and Sell Stock | 968. BT Camera
Day 38 | 509. Fibonacci Number | 70. Climbing Stairs | 746. Min Cost Climbing Stairs
Day 39 | 62. Unique Paths | 63. Unique Paths II
Day 41 | 343. Integer Break | 96. Unique Binary Search Trees
Day 42 | 0-1 Backpack Basic Theory（一）| 0-1 Backpack Basic Theory（二）| 416. Partition Equal Subset Sum
Day 43 | 1049. Last Stone Weight II | 494. Target Sum | 474. Ones and Zeroes
Day 44 | Full Backpack Basic Theory | 518. Coin Change II | 377. Combination Sum IV
Day 45 | 70. Climbing Stairs | 322. Coin Change | 279. Perfect Squares
Day 46 | 139. Word Break | Backpack Question Summary
Day 48 | 198. House Robber | 213. House Robber II | 337. House Robber III
Day 49 | 121. Best Time to Buy and Sell Stock I | 122. Best Time to Buy and Sell Stock II

---

123. Best Time to Buy and Sell Stock III

Question Link

class Solution {
    public int maxProfit(int[] prices) {
        int[][] dp = new int[prices.length][5];
        dp[0][1] = -prices[0];
        dp[0][3] = -prices[0];
        for(int i = 1; i < prices.length; i++){
            dp[i][1] = Math.max(dp[i-1][1], dp[i-1][0] - prices[i]);
            dp[i][2] = Math.max(dp[i-1][2], dp[i-1][1] + prices[i]);
            dp[i][3] = Math.max(dp[i-1][3], dp[i-1][2] - prices[i]);
            dp[i][4] = Math.max(dp[i-1][4], dp[i-1][3] + prices[i]);
        }
        return dp[prices.length-1][4];
    }
}

• Meaning of dp array
  • dp[i][0]: no operation(In fact, we can also not set this state)
  • dp[i][1]: maximum cash of first holding the stock on day i
  • dp[i][2]: maximum cash of first doesn’t holding stocks on day i
  • dp[i][3]: maximum cash of second holding the stock on the day i
  • dp[i][4]: maximum cash of second holding the stock on the day i
  • The hold here doesn’t mean buy today. It is also possible that I bought it yesterday and keep it today.
• Recursion Formula:
  • If hold the stock on i-1 day, let’s keep it that way.
  • If first buy stock on day i, current cash is dp[i - 1][0] - prices[i]
  • So the dp[i][1] should choose the larger one: dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] - prices[i])
  • If doesn’t hold the the stock on i - 1, let’s keep it that way.
  • If first sell stock on day i, current cash is prices[i] + dp[i - 1][1]
  • Again dp[i][2] takes the larger one: dp[i][2] = max(dp[i - 1][2], dp[i - 1][1] + prices[i])
  • So dp[i][3] and dp[i][4] are the similar like above. The complete recursion formulas are as follows.
    • dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0] - prices[i])
    • dp[i][2] = Math.max(dp[i - 1][2], dp[i - 1][1] + prices[i])
    • dp[i][3] = Math.max(dp[i - 1][3], dp[i - 1][2] - prices[i])
    • dp[i][4] = Math.max(dp[i - 1][4], dp[i - 1][3] + prices[i])
• Initialize the DP array
  • According to the recursion formula, the values of dp array are derived from dp[0][0], dp[0][1], dp[0][2], dp[0][3] and dp[0][4]
  • dp[0][0] = 0: no operation
  • dp[0][1] = -prices[0]: buy the stock on day 0
  • dp[0][2] = 0: first sell the stock on day 0
  • dp[0][3] = -prices[0]: second buy the stock on the day 0
  • dp[0][4]= 0: second sell the stock on the day 0
• Traversal Order is from front to back.

188. Best Time to Buy and Sell Stock IV

Question Link

class Solution {
    public int maxProfit(int k, int[] prices) {
        int[][] dp = new int[prices.length][2*k+1];
        for(int m = 0; m < k; m++)
            dp[0][2*m+1] = -prices[0];
        
        for(int i = 1; i < prices.length; i++){
            for(int m = 1; m <= k; m++){
                // buy
                dp[i][2*m-1] = Math.max(dp[i-1][2*m-1], dp[i-1][2*m-2] - prices[i]);
                // sell
                dp[i][2*m] = Math.max(dp[i-1][2*m], dp[i-1][2*m-1] + prices[i]);
            }
        }
        return dp[prices.length-1][2*k];
    }
}

• Compare this question to the 123. Best Time to Buy and Sell Stock III, the most significant difference is buying the stock when k is odd and selling the stock when k is even.