Day 57 | 647. Palindromic Substrings | 516. Longest Palindromic Subsequence

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
Day 50 | 123. Best Time to Buy and Sell Stock III | 188. Best Time to Buy and Sell Stock IV
Day 51 | 309. Best Time to Buy and Sell Stock with Cooldown | 714. with Transaction Fee
Day 52 | 300. Longest Increasing Subsequence | 674. Longest Continuous Increasing Subsequence | 718. Maximum Length of Repeated Subarray
Day 53 | 1143. Longest Common Subsequence | 1035. Uncrossed Lines | 53. Maximum Subarray
Day 55 | 392. Is Subsequence | 115. Distinct Subsequences
Day 56 | 583. Delete Operation for Two Strings | 72. Edit Distance

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647. Palindromic Substrings

Question Link

class Solution {
    public int countSubstrings(String s) {
        char[] charS = s.toCharArray();
        int len = s.length();
        boolean[][] dp = new boolean[len][len];
        int result = 0;
        for(int i = len-1; i >= 0; i--){
            for(int j = i; j < len; j++){
                if(charS[i] == charS[j]){
                    if(j-i <= 1){               // case 1,2
                        dp[i][j] = true;
                        result++;
                    }else if(dp[i+1][j-1]){     // case 3
                        dp[i][j] = true;
                        result++;
                    }
                }
            }
        }
        return result;
    }
}

• dp[i][j]: whether a substring in the range [i, j] is a palindrome substring.
• Recursive Formula
  • If s[i] is different with s[j], dp[i][j] must be false.
  • If s[i] is the same as s[j], there are three cases as follows:
    • Same index like i == j, s[i] and s[j] are the same character
    • The difference between i and j is 1, like aa. It is also a palindromic.
    • The difference between i and j is greater than 1, like cabac. At this time, s[i] and s[j] are already the same. Whether dp[i, j] is palindromic depends on whether dp[i+1, j-1] is palindromic.
• All elements in the array should be initialized to false.
• Traversal Order: According to the case 3 of recursive formula, dp[i][j] is derived from dp[i+1][j-1]. So the traversal order must be bottom to top, left to right.

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516. Longest Palindromic Subsequence

Question Link

class Solution {
    public int longestPalindromeSubseq(String s) {  
        char[] charS = s.toCharArray();
        int[][] dp = new int[charS.length][charS.length];
        for(int i = 0; i < charS.length; i++) dp[i][i] = 1;

        for(int i = charS.length-1; i >= 0; i--){
            for(int j = i+1; j < charS.length; j    ++){
                if(charS[i] == charS[j])
                    dp[i][j] = dp[i+1][j-1] + 2;
                else
                    dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1]);
            }
        }
        return dp[0][charS.length-1];
    }
}

• The palindromic substring is continuous, palindromic subsequence is not continuous.
• dp[i][j]: the length of the longest palindrome subsequence in the range [i, j].
• Recursive Formula
  • If s[i] is the same as s[j]
    • dp[i][j] = dp[i+1][j-1] + 2
  • If s[i] is different with s[j]
    • The length of palindromic subsequence adding s[i] is dp[i][j-1]
    • The length of palindromic subsequence adding s[j] is dp[i+1][j]
    • Choose the larger one: dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])
• Array Initialization
  • According to the dp[i][j] = dp[i+1][j-1] + 2, the recursive formula won’t be able to calculate the dp[i][j] when i is the same as j.
  • So dp[i][j](when i == j) must be initialized to 1.
• Traversal Order
  • dp[i][j] is derived from dp[i + 1][j - 1]，dp[i + 1][j] and dp[i][j - 1].
  • So we must traversal order must be bottom to top, left to right.
  • Since dp[i][j] has already been initialized to 1 when i is the same as j. We must start from i+1 to traverse j.