Saturday, October 15, 2016

LeetCode 420 - Strong Password Checker


https://leetcode.com/problems/strong-password-checker/
A password is considered strong if below conditions are all met:
  1. It has at least 6 characters and at most 20 characters.
  2. It must contain at least one lowercase letter, at least one uppercase letter, and at least one digit.
  3. It must NOT contain three repeating characters in a row ("...aaa..." is weak, but "...aa...a..." is strong, assuming other conditions are met).
Write a function strongPasswordChecker(s), that takes a string s as input, and return the MINIMUM change required to make s a strong password. If s is already strong, return 0.
Insertion, deletion and replace of any one character are all considered as one change.

https://discuss.leetcode.com/topic/63185/java-easy-solution-with-explanation
    public int strongPasswordChecker(String s) {
        
        if(s.length()<2) return 6-s.length();
        
        //Initialize the states, including current ending character(end), existence of lowercase letter(lower), uppercase letter(upper), digit(digit) and number of replicates for ending character(end_rep)
        char end = s.charAt(0);
        boolean upper = end>='A'&&end<='Z', lower = end>='a'&&end<='z', digit = end>='0'&&end<='9';
        
        //Also initialize the number of modification for repeated characters, total number needed for eliminate all consequnce 3 same character by replacement(change), and potential maximun operation of deleting characters(delete). Note delete[0] means maximum number of reduce 1 replacement operation by 1 deletion operation, delete[1] means maximun number of reduce 1 replacement by 2 deletion operation, delete[2] is no use here. 
        int end_rep = 1, change = 0;
        int[] delete = new int[3];
        
        for(int i = 1;i<s.length();++i){
            if(s.charAt(i)==end) ++end_rep;
            else{
                change+=end_rep/3;
                if(end_rep/3>0) ++delete[end_rep%3];
                //updating the states
                end = s.charAt(i);
                upper = upper||end>='A'&&end<='Z';
                lower = lower||end>='a'&&end<='z';
                digit = digit||end>='0'&&end<='9';
                end_rep = 1;
            }
        }
        change+=end_rep/3;
        if(end_rep/3>0) ++delete[end_rep%3];
        
        //The number of replcement needed for missing of specific character(lower/upper/digit)
        int check_req = (upper?0:1)+(lower?0:1)+(digit?0:1);
        
        if(s.length()>20){
            int del = s.length()-20;
            
            //Reduce the number of replacement operation by deletion
            if(del<=delete[0]) change-=del;
            else if(del-delete[0]<=2*delete[1]) change-=delete[0]+(del-delete[0])/2;
            else change-=delete[0]+delete[1]+(del-delete[0]-2*delete[1])/3;
            
            return del+Math.max(check_req,change);
        }
        else return Math.max(6-s.length(), Math.max(check_req, change));
    }
http://www.cnblogs.com/zslhq/p/5986623.html
http://www.cnblogs.com/grandyang/p/5988792.html
https://discuss.leetcode.com/topic/63854/o-n-java-solution-by-analyzing-changes-allowed-to-fix-each-condition
The basic principle is straightforward: if we want to make MINIMUM change to turn s into a strong password, each change made should fix as many problems as possible.
So first let's figure out what changes are suitable for righting each condition. Just to clarify, each change here should be characterized by at least two factors: the type of operation it takes and the position in the string where the operation is applied.
(Note: Ideally we should also include the characters involved in the operation and the "power" of each operation for eliminating problems but they turn out to be partially relevant so I will mention them only when appropriate.)
  1. Length problem: if the total length is less than 6, the change that should be made is (insert, any position), which reads as "the operation is insertion and it can be applied to anywhere in the string". If the total length is greater than 20, then the change should be (delete, any position).
  2. Missing letter or digit: if any of the lowercase/uppercase letters or digits is missing, we can do either (insert, any position) or (replace, any position) to correct it. (Note here the characters for insertion or replacement can only be those missing.)
  3. Repeating characters: for repeating characters, all three operations are allowed but the positions where they can be applied are limited within the repeating characters. For example, to fix "aaaaa", we can do one replacement (replace the middle 'a') or two insertion (one after the second 'a' and one after the fourth 'a') or three deletion (delete any of the three 'a's). So the possible changes are (replace, repeating characters), (insert, repeating characters), (delete, repeating characters). (Note here the "power" of each operation for fixing the problem are different -- replacement is the strongest while deletion is the weakest.)
All right, what's next? If we want a change to eliminate as many problems as it can, it must be shared among the possible solutions to each problem it can fix. So our task is to find out possible overlappings among the changes for fixing each problem.
Since there are most (three) changes allowed for the third condition, we may start from combinations first condition & third conditionand second condition & third condition. It's not too hard to conclude that any change that can fix the first condition or second condition is also able to fix the third one (since the type of operation here is irrelevant, we are free to choose the position of the operation to match those of the repeating characters). For combination first condition & second condition, depending on the length of the string, there will be overlapping if length is less than 6 or no overlapping if length is greater than 20.
From the analyses above, it seems worthwhile to distinguish between the two cases: when the string length is too short or too long.
For the former case, it can be shown that the changes needed to fix the first and second conditions always outnumber those for the third one. Since whatever change used fixing the first two conditions can also correct the third one, we may concern ourselves with only the first two conditions. Also as there are overlappings between the changes for fixing these two conditions, we will prefer those overlapping ones, i.e. (insert, any position). Another point is that the characters involved in the operation matters now. To fix the first condition, only those missing characters can be inserted while for the second condition, it can be any character. Therefore correcting the first condition takes precedence over the second one.
For the latter case, there are overlappings between the first & third and second & third conditions, so those changes will be taken, i.e., first condition => (delete, any position), second condition => (replace, any position). The reason not to use (insert, any position) for the second condition is that it contradicts the changes made to the first condition (therefore has the tendency to cancel its effects). After fixing the first two conditions, what operations should we choose for the third one?
Now the "power" of each operation for eliminating problems comes into play. For the third condition, the "power" of each operation will be measured by the maximum number of repeating characters it is able to get rid of. For example, one replacement can eliminate at most 5 repeating characters while insertion and deletion can do at most 4 and 3, respectively. In this case, we say replacement has more "power" than insertion or deletion. Intuitively the more "powerful" the operation is, the less number of changes is needed for correcting the problem. Therefore (replace, repeating characters) triumphs in terms of fixing the third condition.
Further more, another very interesting point shows up when the "power" of operation is taken into consideration (And thank yicui for pointing it out). As I mentioned that there are overlappings between changes made for fixing the first two conditions and for the third one, which means the operations chosen above for the first two conditions will also be applied to the third one. For the second condition with change chosen as (replace, any position), we have no problem adapting it so that it coincides with the optimal change (replace, repeating characters) for the third condition. However, there is no way to do that for the first condition with change (delete, any position). We have a conflict now!
So how do we reconcile it? The trick is that for a sequence of repeating characters of length k (k >=3), instead of turning it all the way into a sequence of length 2 (so as to fix the repeating character problem) by the change (delete, any position), we will first reduce its length to (3m + 2), where (3m + 2) is the largest integer of the form yet no more than k. That is to say, if k is a multiple of 3, we apply once such change so its length will become (k - 1); else if k is a multiple of 3 plus 1, we apply twice such change to cut its length down to (k - 2), provided we have more such changes to spare (note here we need at least two changes but the remaining available changes may be less than that, so we should stick to the smaller one: 2 or the remaining available changes). The reason is that the optimal change (replace, repeating characters) for the third condition will be most powerful when the total length of the repeating characters is of this form. Of course, if we still have more changes (delete, any position) to do after that, then we are free to turn the repeating sequence all the way into a sequence of length 2.

A quick explanation of the program:
  1. "res" denotes the minimum changes; "a", "A" and "d" are the number of missing lowercase letters, uppercase letters and digits, respectively; "arr" is an integer array whose element will be the number of repeating characters starting at the corresponding position in the string.
  2. In the following loop we fill in the values for "a", "A", "d" and "arr" to identify the problems for each condition. The total number of missing characters "total_missing" will be the summation of "a", "A", "d" and fixing this problem takes at least "total_missing" changes.
  3. We then distinguish the two cases when the string is too short or too long. If it is too short, we pad its length to at least 6 (note in this case we've already inserted "total_missing" characters so the new length is the summation of the original length and "total_missing").
  4. Otherwise, to fix the first condition, we need to delete "over_len" (number of surplus characters) characters. Since fixing the first condition also corrects the third one, we need to get rid of those parts from the "arr" array. And as I mentioned, we need to first turn all numbers in the "arr" array greater than 2 into the form of (3m + 2) and then reduce them all the way to 2 if "over_len" is still greater than 0. After that, we need to replace "total_missing" characters to fix the second condition, which also fixes part (or all) of the third condition. Therefore we only need to take the larger number of changes needed for fixing the second condition (which is "total_missing") and the third condition (which is "left_over", as it is the number after fixing the first one).
Both time and space complexity is O(n). Not sure if we can reduce the space down to O(1) by computing the "arr" array on the fly.
public int strongPasswordChecker(String s) {
    int res = 0, a = 1, A = 1, d = 1;
    char[] carr = s.toCharArray();
    int[] arr = new int[carr.length];
        
    for (int i = 0; i < arr.length;) {
        if (Character.isLowerCase(carr[i])) a = 0;
        if (Character.isUpperCase(carr[i])) A = 0;
        if (Character.isDigit(carr[i])) d = 0;
            
        int j = i;
        while (i < carr.length && carr[i] == carr[j]) i++;
        arr[j] = i - j;
    }
        
    int total_missing = (a + A + d);

    if (arr.length < 6) {
        res += total_missing + Math.max(0, 6 - (arr.length + total_missing));
            
    } else {
        int over_len = Math.max(arr.length - 20, 0), left_over = 0;
        res += over_len;
            
        for (int k = 1; k < 3; k++) {
            for (int i = 0; i < arr.length && over_len > 0; i++) {
                if (arr[i] < 3 || arr[i] % 3 != (k - 1)) continue;
                arr[i] -= Math.min(over_len, k);
                over_len -= k;
            }
        }
            
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] >= 3 && over_len > 0) {
                int need = arr[i] - 2;
                arr[i] -= over_len;
                over_len -= need;
            }
                
            if (arr[i] >= 3) left_over += arr[i] / 3;
        }
            
        res += Math.max(total_missing, left_over);
    }
        
    return res;
}

https://discuss.leetcode.com/topic/63185/java-easy-solution-with-explanation
The general idea is to record some states, and calculate the edit distance at the end. All detail are explained in the comments.
    public int strongPasswordChecker(String s) {
        
        if(s.length()<2) return 6-s.length();
        
        //Initialize the states, including current ending character(end), existence of lowercase letter(lower), uppercase letter(upper), digit(digit) and number of replicates for ending character(end_rep)
        char end = s.charAt(0);
        boolean upper = end>='A'&&end<='Z', lower = end>='a'&&end<='z', digit = end>='0'&&end<='9';
        
        //Also initialize the number of modification for repeated characters, total number needed for eliminate all consequnce 3 same character by replacement(change), and potential maximun operation of deleting characters(delete). Note delete[0] means maximum number of reduce 1 replacement operation by 1 deletion operation, delete[1] means maximun number of reduce 1 replacement by 2 deletion operation, delete[2] is no use here. 
        int end_rep = 1, change = 0;
        int[] delete = new int[3];
        
        for(int i = 1;i<s.length();++i){
            if(s.charAt(i)==end) ++end_rep;
            else{
                change+=end_rep/3;
                if(end_rep/3>0) ++delete[end_rep%3];
                //updating the states
                end = s.charAt(i);
                upper = upper||end>='A'&&end<='Z';
                lower = lower||end>='a'&&end<='z';
                digit = digit||end>='0'&&end<='9';
                end_rep = 1;
            }
        }
        change+=end_rep/3;
        if(end_rep/3>0) ++delete[end_rep%3];
        
        //The number of replcement needed for missing of specific character(lower/upper/digit)
        int check_req = (upper?0:1)+(lower?0:1)+(digit?0:1);
        
        if(s.length()>20){
            int del = s.length()-20;
            
            //Reduce the number of replacement operation by deletion
            if(del<=delete[0]) change-=del;
            else if(del-delete[0]<=2*delete[1]) change-=delete[0]+(del-delete[0])/2;
            else change-=delete[0]+delete[1]+(del-delete[0]-2*delete[1])/3;
            
            return del+Math.max(check_req,change);
        }
        else return Math.max(6-s.length(), Math.max(check_req, change));
    }
}


No comments:

Post a Comment

Labels

GeeksforGeeks (959) Algorithm (811) LeetCode (632) to-do (596) Classic Algorithm (334) Review (330) Classic Interview (299) Dynamic Programming (263) Google Interview (229) LeetCode - Review (224) Tree (146) POJ (137) Difficult Algorithm (136) EPI (127) Different Solutions (118) Bit Algorithms (110) Cracking Coding Interview (110) Smart Algorithm (109) Math (91) HackerRank (85) Lintcode (83) Binary Search (73) Graph Algorithm (73) Interview Corner (61) Greedy Algorithm (60) List (58) Binary Tree (56) DFS (54) Algorithm Interview (53) Advanced Data Structure (52) Codility (52) ComProGuide (52) LeetCode - Extended (47) USACO (46) Geometry Algorithm (44) BFS (43) Data Structure (42) Mathematical Algorithm (42) ACM-ICPC (41) Interval (38) Jobdu (38) Recursive Algorithm (38) Stack (38) String Algorithm (38) Binary Search Tree (37) Knapsack (37) Codeforces (36) Introduction to Algorithms (36) Matrix (36) Must Known (36) Beauty of Programming (35) Sort (35) Array (33) prismoskills (33) Segment Tree (32) Space Optimization (32) Trie (32) Union-Find (32) Backtracking (31) HDU (31) Google Code Jam (30) Permutation (30) Puzzles (30) Array O(N) (29) Data Structure Design (29) Company-Zenefits (28) Microsoft 100 - July (28) to-do-must (28) Random (27) Sliding Window (26) GeeksQuiz (25) Logic Thinking (25) hihocoder (25) High Frequency (23) Palindrome (23) Algorithm Game (22) Company - LinkedIn (22) Graph (22) Queue (22) DFS + Review (21) Hash (21) TopCoder (21) Binary Indexed Trees (20) Brain Teaser (20) CareerCup (20) Company - Twitter (20) Pre-Sort (20) Company-Facebook (19) UVA (19) Probabilities (18) Follow Up (17) Codercareer (16) Company-Uber (16) Game Theory (16) Heap (16) Shortest Path (16) String Search (16) Topological Sort (16) Tree Traversal (16) itint5 (16) Iterator (15) Merge Sort (15) O(N) (15) Difficult (14) Number (14) Number Theory (14) Post-Order Traverse (14) Priority Quieue (14) Amazon Interview (13) BST (13) Basic Algorithm (13) Bisection Method (13) Codechef (13) Majority (13) mitbbs (13) Combination (12) Computational Geometry (12) KMP (12) Long Increasing Sequence(LIS) (12) Modify Tree (12) Reconstruct Tree (12) Reservoir Sampling (12) 尺取法 (12) AOJ (11) DFS+Backtracking (11) Fast Power Algorithm (11) Graph DFS (11) LCA (11) LeetCode - DFS (11) Ordered Stack (11) Princeton (11) Tree DP (11) 挑战程序设计竞赛 (11) Binary Search - Bisection (10) Company - Microsoft (10) Company-Airbnb (10) Euclidean GCD (10) Facebook Hacker Cup (10) HackerRank Easy (10) Reverse Thinking (10) Rolling Hash (10) SPOJ (10) Theory (10) Tutorialhorizon (10) X Sum (10) Coin Change (9) Lintcode - Review (9) Mathblog (9) Max-Min Flow (9) Stack Overflow (9) Stock (9) Two Pointers (9) Book Notes (8) Bottom-Up (8) DP-Space Optimization (8) Divide and Conquer (8) Graph BFS (8) LeetCode - DP (8) LeetCode Hard (8) Prefix Sum (8) Prime (8) System Design (8) Tech-Queries (8) Use XOR (8) 穷竭搜索 (8) Algorithm Problem List (7) DFS+BFS (7) Facebook Interview (7) Fibonacci Numbers (7) Game Nim (7) HackerRank Difficult (7) Hackerearth (7) Interval Tree (7) Linked List (7) Longest Common Subsequence(LCS) (7) Math-Divisible (7) Miscs (7) O(1) Space (7) Probability DP (7) Radix Sort (7) Simulation (7) Suffix Tree (7) Xpost (7) n00tc0d3r (7) 蓝桥杯 (7) Bucket Sort (6) Catalan Number (6) Classic Data Structure Impl (6) DFS+DP (6) DP - Tree (6) How To (6) Interviewstreet (6) Knapsack - MultiplePack (6) Level Order Traversal (6) Manacher (6) Minimum Spanning Tree (6) One Pass (6) Programming Pearls (6) Quick Select (6) Rabin-Karp (6) Randomized Algorithms (6) Sampling (6) Schedule (6) Suffix Array (6) Threaded (6) Time Complexity (6) reddit (6) AI (5) Art Of Programming-July (5) Big Data (5) Brute Force (5) Code Kata (5) Codility-lessons (5) Coding (5) Company - WMware (5) Crazyforcode (5) DFS+Cache (5) DP-Multiple Relation (5) DP-Print Solution (5) Dutch Flag (5) Fast Slow Pointers (5) Graph Cycle (5) Hash Strategy (5) Immutability (5) Inversion (5) Java (5) Kadane - Extended (5) Kadane’s Algorithm (5) Matrix Chain Multiplication (5) Microsoft Interview (5) Morris Traversal (5) Pruning (5) Quadtrees (5) Quick Partition (5) Quora (5) SPFA(Shortest Path Faster Algorithm) (5) Subarray Sum (5) Traversal Once (5) TreeMap (5) jiuzhang (5) to-do-2 (5) 单调栈 (5) 树形DP (5) 1point3acres (4) Anagram (4) Approximate Algorithm (4) Backtracking-Include vs Exclude (4) Brute Force - Enumeration (4) Chess Game (4) Company-Amazon (4) Consistent Hash (4) Convex Hull (4) Cycle (4) DP-Include vs Exclude (4) Dijkstra (4) Distributed (4) Eulerian Cycle (4) Flood fill (4) Graph-Classic (4) HackerRank AI (4) Histogram (4) Kadane Max Sum (4) Knapsack - Mixed (4) Knapsack - Unbounded (4) Left and Right Array (4) MinMax (4) Multiple Data Structures (4) N Queens (4) Nerd Paradise (4) Parallel Algorithm (4) Practical Algorithm (4) Pre-Sum (4) Probability (4) Programcreek (4) Quick Sort (4) Spell Checker (4) Stock Maximize (4) Subsets (4) Sudoku (4) Sweep Line (4) Symbol Table (4) TreeSet (4) Triangle (4) Water Jug (4) Word Ladder (4) algnotes (4) fgdsb (4) 最大化最小值 (4) A Star (3) Abbreviation (3) Algorithm - Brain Teaser (3) Algorithm Design (3) Anagrams (3) B Tree (3) Big Data Algorithm (3) Binary Search - Smart (3) Caterpillar Method (3) Coins (3) Company - Groupon (3) Company - Indeed (3) Cumulative Sum (3) DP-Fill by Length (3) DP-Two Variables (3) Dedup (3) Dequeue (3) Dropbox (3) Easy (3) Edit Distance (3) Expression (3) Finite Automata (3) Forward && Backward Scan (3) Github (3) GoLang (3) Include vs Exclude (3) Joseph (3) Jump Game (3) Knapsack-多重背包 (3) LeetCode - Bit (3) LeetCode - TODO (3) Linked List Merge Sort (3) LogN (3) Maze (3) Min Cost Flow (3) Minesweeper (3) Missing Numbers (3) NP Hard (3) Online Algorithm (3) Pascal's Triangle (3) Pattern Match (3) Project Euler (3) Rectangle (3) Scala (3) SegmentFault (3) Stack - Smart (3) State Machine (3) Streaming Algorithm (3) Subset Sum (3) Subtree (3) Transform Tree (3) Two Pointers Window (3) Warshall Floyd (3) With Random Pointer (3) Word Search (3) bookkeeping (3) codebytes (3) Activity Selection Problem (2) Advanced Algorithm (2) AnAlgorithmADay (2) Application of Algorithm (2) Array Merge (2) BOJ (2) BT - Path Sum (2) Balanced Binary Search Tree (2) Bellman Ford (2) Binomial Coefficient (2) Bit Mask (2) Bit-Difficult (2) Bloom Filter (2) Book Coding Interview (2) Branch and Bound Method (2) Clock (2) Codesays (2) Company - Baidu (2) Complete Binary Tree (2) DFS+BFS, Flood Fill (2) DP - DFS (2) DP-3D Table (2) DP-Classical (2) DP-Output Solution (2) DP-Slide Window Gap (2) DP-i-k-j (2) DP-树形 (2) Distributed Algorithms (2) Divide and Conqure (2) Doubly Linked List (2) GoHired (2) Graham Scan (2) Graph - Bipartite (2) Graph BFS+DFS (2) Graph Coloring (2) Graph-Cut Vertices (2) Hamiltonian Cycle (2) Huffman Tree (2) In-order Traverse (2) Include or Exclude Last Element (2) Information Retrieval (2) Interview - Linkedin (2) Invariant (2) Islands (2) Knuth Shuffle (2) LeetCode - Recursive (2) Linked Interview (2) Linked List Sort (2) Longest SubArray (2) Lucene-Solr (2) MST (2) MST-Kruskal (2) Master Theorem (2) Math-Remainder Queue (2) Matrix Power (2) Minimum Vertex Cover (2) Negative All Values (2) Number Each Digit (2) Numerical Method (2) Object Design (2) Order Statistic Tree (2) Palindromic (2) Parentheses (2) Parser (2) Peak (2) Programming (2) Range Minimum Query (2) Reuse Forward Backward (2) Robot (2) Rosettacode (2) Scan from right (2) Search (2) Shuffle (2) Sieve of Eratosthenes (2) SimHash (2) Simple Algorithm (2) Skyline (2) Spatial Index (2) Stream (2) Strongly Connected Components (2) Summary (2) TV (2) Tile (2) Traversal From End (2) Tree Sum (2) Tree Traversal Return Multiple Values (2) Word Break (2) Word Graph (2) Word Trie (2) Young Tableau (2) 剑指Offer (2) 数位DP (2) 1-X (1) 51Nod (1) Akka (1) Algorithm - How To (1) Algorithm - New (1) Algorithm Series (1) Algorithms Part I (1) Analysis of Algorithm (1) Array-Element Index Negative (1) Array-Rearrange (1) Auxiliary Array (1) Auxiliary Array: Inc&Dec (1) BACK (1) BK-Tree (1) BZOJ (1) Basic (1) Bayes (1) Beauty of Math (1) Big Integer (1) Big Number (1) Binary (1) Binary Tree Variant (1) Bipartite (1) Bit-Missing Number (1) BitMap (1) BitMap index (1) BitSet (1) Bug Free Code (1) BuildIt (1) C/C++ (1) CC Interview (1) Cache (1) Calculate Height at Same Recusrion (1) Cartesian tree (1) Check Tree Property (1) Chinese (1) Circular Buffer (1) Code Quality (1) Codesolutiony (1) Company - Alibaba (1) Company - Palantir (1) Company - WalmartLabs (1) Company-Apple (1) Company-Epic (1) Company-Salesforce (1) Company-Snapchat (1) Company-Yelp (1) Compression Algorithm (1) Concurrency (1) Convert BST to DLL (1) Convert DLL to BST (1) Custom Sort (1) Cyclic Replacement (1) DFS-Matrix (1) DP - Probability (1) DP Fill Diagonal First (1) DP-Difficult (1) DP-End with 0 or 1 (1) DP-Fill Diagonal First (1) DP-Graph (1) DP-Left and Right Array (1) DP-MaxMin (1) DP-Memoization (1) DP-Node All Possibilities (1) DP-Optimization (1) DP-Preserve Previous Value (1) DP-Print All Solution (1) Database (1) Detect Negative Cycle (1) Directed Graph (1) Do Two Things at Same Recusrion (1) Domino (1) Dr Dobb's (1) Duplicate (1) Equal probability (1) External Sort (1) FST (1) Failure Function (1) Fraction (1) Front End Pointers (1) Funny (1) Fuzzy String Search (1) Game (1) Generating Function (1) Generation (1) Genetic algorithm (1) GeoHash (1) Geometry - Orientation (1) Google APAC (1) Graph But No Graph (1) Graph Transpose (1) Graph Traversal (1) Graph-Coloring (1) Graph-Longest Path (1) Gray Code (1) HOJ (1) Hanoi (1) Hard Algorithm (1) How Hash (1) How to Test (1) Improve It (1) In Place (1) Inorder-Reverse Inorder Traverse Simultaneously (1) Interpolation search (1) Interview (1) Interview - Easy (1) Interview - Facebook (1) Isomorphic (1) JDK8 (1) K Dimensional Tree (1) Knapsack - Fractional (1) Knapsack - ZeroOnePack (1) Knight (1) Kosaraju’s algorithm (1) Kruskal (1) Kruskal MST (1) Kth Element (1) Least Common Ancestor (1) LeetCode - Binary Tree (1) LeetCode - Coding (1) LeetCode - Detail (1) LeetCode - Related (1) LeetCode Diffcult (1) Linked List Reverse (1) Linkedin (1) Linkedin Interview (1) Local MinMax (1) Logic Pattern (1) Longest Common Subsequence (1) Longest Common Substring (1) Longest Prefix Suffix(LPS) (1) Manhattan Distance (1) Map && Reverse Map (1) Math - Induction (1) Math-Multiply (1) Math-Sum Of Digits (1) Matrix - O(N+M) (1) Matrix BFS (1) Matrix Graph (1) Matrix Search (1) Matrix+DP (1) Matrix-Rotate (1) Max Min So Far (1) Median (1) Memory-Efficient (1) MinHash (1) MinMax Heap (1) Monotone Queue (1) Monto Carlo (1) Multi-Reverse (1) Multiple DFS (1) Multiple Tasks (1) Next Successor (1) Offline Algorithm (1) PAT (1) Partition (1) Path Finding (1) Patience Sort (1) Persistent (1) Pigeon Hole Principle (1) Power Set (1) Pratical Algorithm (1) Probabilistic Data Structure (1) Proof (1) Python (1) Queue & Stack (1) RSA (1) Ranking (1) Rddles (1) ReHash (1) Realtime (1) Recurrence Relation (1) Recursive DFS (1) Recursive to Iterative (1) Red-Black Tree (1) Region (1) Regular Expression (1) Resources (1) Reverse Inorder Traversal (1) Robin (1) Selection (1) Self Balancing BST (1) Similarity (1) Sort && Binary Search (1) String Algorithm. Symbol Table (1) String DP (1) String Distance (1) SubMatrix (1) Subsequence (1) System of Difference Constraints(差分约束系统) (1) TSP (1) Ternary Search Tree (1) Test (1) Thread (1) TimSort (1) Top-Down (1) Tournament (1) Tournament Tree (1) Transform Tree in Place (1) Tree Diameter (1) Tree Rotate (1) Trie + DFS (1) Trie and Heap (1) Trie vs Hash (1) Trie vs HashMap (1) Triplet (1) Two Data Structures (1) Two Stacks (1) USACO - Classical (1) USACO - Problems (1) UyHiP (1) Valid Tree (1) Vector (1) Wiggle Sort (1) Wikipedia (1) Yahoo Interview (1) ZOJ (1) baozitraining (1) codevs (1) cos126 (1) javabeat (1) jum (1) namic Programming (1) sqrt(N) (1) 两次dijkstra (1) 九度 (1) 二进制枚举 (1) 夹逼法 (1) 归一化 (1) 折半枚举 (1) 枚举 (1) 状态压缩DP (1) 男人八题 (1) 英雄会 (1) 逆向思维 (1)

Popular Posts