Day 37 | 738. Monotone Increasing Digits | 714. Best Time to Buy and Sell Stock | 968. BT Camera

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

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LeetCode 738. Monotone Increasing Digits

Question Link

class Solution {
    public int monotoneIncreasingDigits(int n) {
        String  s = String.valueOf(n);
        char[] chars = s.toCharArray();
        int start = s.length();
        for(int i = s.length() - 2; i >= 0; i--){
            if(chars[i] > chars[i+1]){
                chars[i]--;
                start = i + 1;
            }
        }

        for(int i = start; i < s.length(); i++)
            chars[i] = '9';
        
        return Integer.parseInt(String.valueOf(chars));
    }
}

• When meeting chars[i - 1] > chars[i], let chars[i-1]-- and set chars[i] as 9. That can ensure these two digits become the largest monotone-increasing integers
• The traverse order is from back to front. Because if we traverse from front to back, we can’t reuse the result of the last compare

LeetCode 714. Best Time to Buy and Sell Stock with Transaction Fee

Question Link

class Solution {
    public int maxProfit(int[] prices, int fee) {
        int buy = prices[0] + fee;
        int sum = 0;
        for(int p : prices){
            if(p + fee < buy)
                buy = p + fee;
            
            else if(p > buy){
                sum += p - buy;
                buy = p;
            }
        }
        return sum;
    }
}

LeetCode 968. BT Camera

Question Link

class Solution {
    int res = 0;
    public int minCameraCover(TreeNode root) {
        if(traverse(root) == 0)
            res++;
        return res;
    }

    int traverse(TreeNode root){
        if(root == null)
            return 2;   // By default, empty nodes have coverage to avoid install cameras on the leaf
        int left = traverse(root.left);
        int right = traverse(root.right);

        if(left == 2 && right == 2)
            return 0;
        else if(left == 0 || right == 0){
            res++;
            return 1;
        }else
            return 2;
    }   
}

• The status of nodes:
  • 0 means no coverage
  • 1 means the node has the camera
  • 2 means the node has coverage
• By default, empty nodes have coverage to avoid installing cameras on the leaf