Day 56 | 583. Delete Operation for Two Strings | 72. Edit Distance

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

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583. Delete Operation for Two Strings

Question Link

Solution 1

class Solution {
    public int minDistance(String word1, String word2) {
        char[] char1 = word1.toCharArray();
        char[] char2 = word2.toCharArray();
        int[][] dp = new int[char1.length+1][char2.length+1];
        int len = 0; // length of longest common subsequence
        for(int i = 1; i <= char1.length; i++){
            for(int j = 1; j <= char2.length; j++){
                if(char1[i-1] == char2[j-1])
                    dp[i][j] = dp[i-1][j-1] + 1;
                else
                    dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
                len = Math.max(len, dp[i][j]);
            }
        }
        return char1.length + char2.length - len*2;
    }
}

• This solution stores the length of longest common subsequence on the dp array.
• We could calculate the length of longest common subsequence first. Then the minimum number of steps is as follows:
  • word1.length() + word2.length() - len * 2

Solution 2

class Solution {
    public int minDistance(String word1, String word2) {
        char[] char1 = word1.toCharArray();
        char[] char2 = word2.toCharArray();
        int[][] dp = new int[char1.length+1][char2.length+1];
        for(int i = 0; i <= char1.length; i++) dp[i][0] = i;
        for(int j = 0; j <= char2.length; j++) dp[0][j] = j;
        
        for(int i = 1; i <= char1.length; i++){
            for(int j = 1; j <= char2.length; j++){
                if(char1[i-1] == char2[j-1])
                    dp[i][j] = dp[i-1][j-1];
                else
                    dp[i][j] = Math.min(dp[i-1][j] + 1, dp[i][j-1] + 1);
            }
        }
        return dp[char1.length][char2.length];
    }
}

• This solution stores the number of character that need to delete on the dp array.
• dp[i][j]: The minimum number of delete operations required to make word1 ends with i-1 and word2 ends with j-1 the same.
• Recursive Formula
  • When word1[i-1] is the same as word2[j-1]:
    • dp[i][j] = dp[i-1][j-1]
  • When word1[i-1] is different from word2[j-1]:
    • Delete word1[i-1], minimum number of operations: dp[i][j] = dp[i-1][j] + 1
    • Delete word2[j-1], minimum number of operations: dp[i][j] = dp[i][j-1] + 1
    • Delete word1[i-1] and word2[j-1], minimum number of operations: dp[i][j] = dp[i-1][j-1] + 2. We do not consider this case, cause dp[i][j-1] already disregarded word2[j - 1]. So when we delete word1[i-1], we can achieve the effect of deleting both elements. That is dp[i][j-1] + 1.
• Array Initialization
  • dp[i][0] = i, cause any word1 have to delete i elements to be a empty string.
  • dp[0][j] = j, cause any word2 have to delete j elements to be a empty string.

72. Edit Distance

Question Link

class Solution {
    public int minDistance(String word1, String word2) {
        char[] char1 = word1.toCharArray();
        char[] char2 = word2.toCharArray();
        int[][] dp = new int[char1.length + 1][char2.length + 1];

        for(int i = 1 ; i <= char1.length; i++) dp[i][0] = i;
        for(int j = 1; j <= char2.length; j++) dp[0][j] = j;

        for(int i = 1 ; i <= char1.length; i++){
            for(int j = 1; j <= char2.length; j++){
                if(char1[i-1] == char2[j-1])
                    dp[i][j] = dp[i-1][j-1];
                else
                    dp[i][j] = Math.min(dp[i-1][j-1], Math.min(dp[i-1][j], dp[i][j-1])) + 1;
            }
        }
        return dp[char1.length][char2.length];
    }
}

• dp[i][j]: The minimum number of operations required to convert word1 ends with i-1 to word2 ends with j-1
• Recursive Formula
  • When word1[i-1] is the same as word2[j-1]:
    • dp[i][j] = dp[i-1][j-1]
  • When word1[i-1] is different from word2[j-1]:
    • Delete/Insertword1[i-1], minimum number of operations: dp[i][j] = dp[i-1][j] + 1.
    • Delete/Insert word2[j-1], minimum number of operations: dp[i][j] = dp[i][j-1] + 1.
    • Replace word1[i-1] with word2[j-1], minimum number of operations: dp[i][j] = dp[i-1][j-1] + 1.
    • Chose the minimum one: dp[i][j] = Math.min(dp[i-1][j-1], Math.min(dp[i-1][j], dp[i][j-1])) + 1.
• Array Initialization
  • dp[i][0] = i, cause any word1 have to delete i elements to be an empty string or word2 has to insert i elements to be the same as word1
  • dp[0][j] = j, cause any word2 have to delete j elements to be an empty string or word1 has to insert j elements to be the same as word2