Day 35 | 860. Lemonade Change | 406. Queue Reconstruction by Height | 452. Minimum Number of Arrows

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

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LeetCode 860. Lemonade Change

Question Link

class Solution {
    public boolean lemonadeChange(int[] bills) {
        int five = 0, ten = 0, twenty = 0;
        for(int bill : bills){
            // 
            if(bill == 5)
                five++;
            else if(bill == 10){
                ten++;
                five--;
            }
            else if(bill == 20){
                if(ten > 0){
                    ten--;
                    five--;
                }
                else
                    five -= 3;
            }
            if(five < 0 || ten < 0)
                return false;
        }
        return true;
    }
}

This question has three situations:

• Bill is $5, receive directly
• Bill is $10. Consume $5, receive $10
• Bill is $20. Prioritize consuming $10 and $5. If $10 is not enough, you can consume three $5.

The third situation is the greedy strategy. Because $10 only can give change for $20. But $5 can give change for $10 and $20. So $5 is more almighty.
Partial Priority：Prioritize consuming $10 when meeting $20 to complete the change.
Global Priority：Complete the change of all bills.

LeetCode 406. Queue Reconstruction by Height

Question Link

class Solution {
    public int[][] reconstructQueue(int[][] people) {
        // Order by height desc. If the height is the same,  the one with a smaller k stands in front
         Arrays.sort(people, (a, b) -> {
            if (a[0] == b[0]) return a[1] - b[1];
            return b[0] - a[0];
        });

        LinkedList<int[]> que = new LinkedList<>();
        for(int[] p : people){
            que.add(p[1], p); // add(index, element)
        }
        return que.toArray(new int[people.length][2]);
    }
}

• Order by height desc. If the height is the same, the one with a smaller k stands in front
• Use k as the index to insert nodes.

LeetCode 452. Minimum Number of Arrows to Burst Balloons

Question Link

class Solution {
    public int findMinArrowShots(int[][] points) {
        // Order by x start ASC
        Arrays.sort(points, (a, b) -> Integer.compare(a[0], b[0]));
        int count = 1;
        for(int i = 1; i < points.length; i++){
            if(points[i][0] > points[i-1][1])   // The two balloons are not next to each other
                count++;
            else                                // The two balloons are next to each other
                points[i][1] = Math.min(points[i][1], points[i-1][1]);      // Update the minimum right border of overlaped balloons
        }

        return count;
    }
}