Monday, June 6, 2016

Find if string is K-Palindrome or not - GeeksforGeeks


Find if string is K-Palindrome or not - GeeksforGeeks
Given a string, find out if the string is K-Palindrome or not. A k-palindrome string transforms into a palindrome on removing at most k characters from it.

If we carefully analyze the problem, the task is to transform the given string into its reverse by removing at most K characters from it. The problem is basically a variation of Edit Distance. We can modify the Edit Distance problem to consider given string and its reverse as input and only operation allowed is deletion. Since given string is compared with its reverse, we will do at most N deletions from first string and N deletions from second string to make them equal. Therefore, for a string to be k-palindrome, 2*N <= 2*K should hold true. Below are the detailed steps of algorithm - Process all characters one by one staring from either from left or right sides of both strings. Let us traverse from the right corner, there are two possibilities for every pair of character being traversed.
  1. If last characters of two strings are same, we ignore last characters and get count for remaining strings. So we recur for lengths m-1 and n-1 where m is length of str1 and n is length of str2.
  2. If last characters are not same, we consider remove operation on last character of first string and last character of second string, recursively compute minimum cost for the operations and take minimum of two values.
    • Remove last char from str1: Recur for m-1 and n.
    • Remove last char from str2: Recur for m and n-1.
int isKPalRec(string str1, string str2, int m, int n)
{
    // If first string is empty, the only option is to
    // remove all characters of second string
    if (m == 0) return n;
    // If second string is empty, the only option is to
    // remove all characters of first string
    if (n == 0) return m;
    // If last characters of two strings are same, ignore
    // last characters and get count for remaining strings.
    if (str1[m-1] == str2[n-1])
        return isKPalRec(str1, str2, m-1, n-1);
    // If last characters are not same,
    // 1. Remove last char from str1 and recur for m-1 and n
    // 2. Remove last char from str2 and recur for m and n-1
    // Take minimum of above two operations
    return 1 + min(isKPalRec(str1, str2, m-1, n), // Remove from str1
                   isKPalRec(str1, str2, m, n-1)); // Remove from str2
}
// Returns true if str is k palindrome.
bool isKPal(string str, int k)
{
    string revStr = str;
    reverse(revStr.begin(), revStr.end());
    int len = str.length();
    return (isKPalRec(str, revStr, len, len) <= k*2);
}

int isKPalDP(string str1, string str2, int m, int n)
{
    // Create a table to store results of subproblems
    int dp[m + 1][n + 1];
    // Fill dp[][] in bottom up manner
    for (int i = 0; i <= m; i++)
    {
        for (int j = 0; j <= n; j++)
        {
            // If first string is empty, only option is to
            // remove all characters of second string
            if (i == 0)
                dp[i][j] = j; // Min. operations = j
            // If second string is empty, only option is to
            // remove all characters of first string
            else if (j == 0)
                dp[i][j] = i; // Min. operations = i
            // If last characters are same, ignore last character
            // and recur for remaining string
            else if (str1[i - 1] == str2[j - 1])
                dp[i][j] = dp[i - 1][j - 1];
            // If last character are different, remove it
            // and find minimum
            else
             dp[i][j] = 1 + min(dp[i - 1][j], // Remove from str1
                             dp[i][j - 1]); // Remove from str2
        }
    }
    return dp[m][n];
}
// Returns true if str is k palindrome.
bool isKPal(string str, int k)
{
    string revStr = str;
    reverse(revStr.begin(), revStr.end());
    int len = str.length();
    return (isKPalDP(str, revStr, len, len) <= k*2);
}
http://www.geeksforgeeks.org/find-if-string-is-k-palindrome-or-not-set-2/
The idea is to find the longest palindromic subsequence of the given string. If the difference between longest palindromic subsequence and the original string is less than equal to k, then the string is k-palindrome else it is not k-palindrome.
For example, longest palindromic subsequence of string abcdeca is acdca(or aceca). The characters which do not contribute to longest palindromic subsequence of the string should be removed in order to make the string palindrome. So on removing b and d (or e) from abcdeca, string will transform into a palindrome.
Longest palindromic subsequence of a string can easily be found using LCS. Following is the two step solution for finding longest palindromic subsequence that uses LCS.
  1. Reverse the given sequence and store the reverse in another array say rev[0..n-1]
  2. LCS of the given sequence and rev[] will be the longest palindromic sequence.
Auxiliary space used by the program is O(n2). It can further be reduced to O(n) by using Space Optimized Solution of LCS.
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs( string X, string Y, int m, int n )
{
    int L[m + 1][n + 1];
    /* Following steps build L[m+1][n+1] in bottom up
        fashion. Note that L[i][j] contains length of
        LCS of X[0..i-1] and Y[0..j-1] */
    for (int i = 0; i <= m; i++)
    {
        for (int j = 0; j <= n; j++)
        {
            if (i == 0 || j == 0)
                L[i][j] = 0;
            else if (X[i - 1] == Y[j - 1])
                L[i][j] = L[i - 1][j - 1] + 1;
            else
                L[i][j] = max(L[i - 1][j], L[i][j - 1]);
        }
    }
    // L[m][n] contains length of LCS for X and Y
    return L[m][n];
}
// find if given string is K-Palindrome or not
bool isKPal(string str, int k)
{
    int n = str.length();
    // Find reverse of string
    string revStr = str;
    reverse(revStr.begin(), revStr.end());
    // find longest palindromic subsequence of
    // given string
    int lps = lcs(str, revStr, n, n);
    // If the difference between longest palindromic
    // subsequence and the original string is less
    // than equal to k, then the string is k-palindrome
    return (n - lps <= k);
}

If we carefully analyze the problem, the task is to transform the given string into its reverse by removing at most K characters from it. The problem is basically a variation of Edit Distance. We can modify the Edit Distance problem to consider given string and its reverse as input and only operation allowed is deletion. Since given string is compared with its reverse, we will do at most N deletions from first string and N deletions from second string to make them equal. Therefore, for a string to be k-palindrome, 2*N <= 2*K should hold true. Below are the detailed steps of algorithm - Process all characters one by one staring from either from left or right sides of both strings. Let us traverse from the right corner, there are two possibilities for every pair of character being traversed.
  1. If last characters of two strings are same, we ignore last characters and get count for remaining strings. So we recur for lengths m-1 and n-1 where m is length of str1 and n is length of str2.
  2. If last characters are not same, we consider remove operation on last character of first string and last character of second string, recursively compute minimum cost for the operations and take minimum of two values.
    • Remove last char from str1: Recur for m-1 and n.
    • Remove last char from str2: Recur for m and n-1.
This problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, re-computations of same subproblems can be avoided by constructing a temporary array that stores results of subproblems .

// find if given string is K-Palindrome or not
int isKPalDP(string str1, string str2, int m, int n)
{
    // Create a table to store results of subproblems
    int dp[m + 1][n + 1];
    // Fill dp[][] in bottom up manner
    for (int i = 0; i <= m; i++)
    {
        for (int j = 0; j <= n; j++)
        {
            // If first string is empty, only option is to
            // remove all characters of second string
            if (i == 0)
                dp[i][j] = j; // Min. operations = j
            // If second string is empty, only option is to
            // remove all characters of first string
            else if (j == 0)
                dp[i][j] = i; // Min. operations = i
            // If last characters are same, ignore last character
            // and recur for remaining string
            else if (str1[i - 1] == str2[j - 1])
                dp[i][j] = dp[i - 1][j - 1];
            // If last character are different, remove it
            // and find minimum
            else
             dp[i][j] = 1 + min(dp[i - 1][j], // Remove from str1
                             dp[i][j - 1]); // Remove from str2
        }
    }
    return dp[m][n];
}
// Returns true if str is k palindrome.
bool isKPal(string str, int k)
{
    string revStr = str;
    reverse(revStr.begin(), revStr.end());
    int len = str.length();
    return (isKPalDP(str, revStr, len, len) <= k*2);
}

If we carefully analyze the problem, the task is to transform the given string into its reverse by removing at most K characters from it. The problem is basically a variation of Edit Distance. We can modify the Edit Distance problem to consider given string and its reverse as input and only operation allowed is deletion. Since given string is compared with its reverse, we will do at most N deletions from first string and N deletions from second string to make them equal. Therefore, for a string to be k-palindrome, 2*N <= 2*K should hold true. Below are the detailed steps of algorithm - Process all characters one by one staring from either from left or right sides of both strings. Let us traverse from the right corner, there are two possibilities for every pair of character being traversed.
  1. If last characters of two strings are same, we ignore last characters and get count for remaining strings. So we recur for lengths m-1 and n-1 where m is length of str1 and n is length of str2.
  2. If last characters are not same, we consider remove operation on last character of first string and last character of second string, recursively compute minimum cost for the operations and take minimum of two values.
    • Remove last char from str1: Recur for m-1 and n.
    • Remove last char from str2: Recur for m and n-1.
// find if given string is K-Palindrome or not
int isKPalRec(string str1, string str2, int m, int n)
{
    // If first string is empty, the only option is to
    // remove all characters of second string
    if (m == 0) return n;
    // If second string is empty, the only option is to
    // remove all characters of first string
    if (n == 0) return m;
    // If last characters of two strings are same, ignore
    // last characters and get count for remaining strings.
    if (str1[m-1] == str2[n-1])
        return isKPalRec(str1, str2, m-1, n-1);
    // If last characters are not same,
    // 1. Remove last char from str1 and recur for m-1 and n
    // 2. Remove last char from str2 and recur for m and n-1
    // Take minimum of above two operations
    return 1 + min(isKPalRec(str1, str2, m-1, n), // Remove from str1
                   isKPalRec(str1, str2, m, n-1)); // Remove from str2
}
// Returns true if str is k palindrome.
bool isKPal(string str, int k)
{
    string revStr = str;
    reverse(revStr.begin(), revStr.end());
    int len = str.length();
    return (isKPalRec(str, revStr, len, len) <= k*2);
}

http://www.geeksforgeeks.org/?p=12998
Dynamic Programming | Set 4 (Longest Common Subsequence)
1) Optimal Substructure: 
Let the input sequences be X[0..m-1] and Y[0..n-1] of lengths m and n respectively. And let L(X[0..m-1], Y[0..n-1]) be the length of LCS of the two sequences X and Y. Following is the recursive definition of L(X[0..m-1], Y[0..n-1]).
If last characters of both sequences match (or X[m-1] == Y[n-1]) then
L(X[0..m-1], Y[0..n-1]) = 1 + L(X[0..m-2], Y[0..n-2])
If last characters of both sequences do not match (or X[m-1] != Y[n-1]) then
L(X[0..m-1], Y[0..n-1]) = MAX ( L(X[0..m-2], Y[0..n-1]), L(X[0..m-1], Y[0..n-2])
  /* Returns length of LCS for X[0..m-1], Y[0..n-1] */
  int lcs( char[] X, char[] Y, int m, int n )
  {
    if (m == 0 || n == 0)
      return 0;
    if (X[m-1] == Y[n-1])
      return 1 + lcs(X, Y, m-1, n-1);
    else
      return max(lcs(X, Y, m, n-1), lcs(X, Y, m-1, n));
  }

  /* Returns length of LCS for X[0..m-1], Y[0..n-1] */
  int lcs( char[] X, char[] Y, int m, int n )
  {
    int L[][] = new int[m+1][n+1];
 
    /* Following steps build L[m+1][n+1] in bottom up fashion. Note
         that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */
    for (int i=0; i<=m; i++)
    {
      for (int j=0; j<=n; j++)
      {
        if (i == 0 || j == 0)
            L[i][j] = 0;
        else if (X[i-1] == Y[j-1])
            L[i][j] = L[i-1][j-1] + 1;
        else
            L[i][j] = max(L[i-1][j], L[i][j-1]);
      }
    }
  return L[m][n];
  }
Read full article from Find if string is K-Palindrome or not - GeeksforGeeks

No comments:

Post a Comment

Labels

GeeksforGeeks (1107) LeetCode (936) Algorithm (805) Review (699) to-do (622) LeetCode - Review (449) Classic Algorithm (330) Classic Interview (296) Dynamic Programming (290) Google Interview (241) Tree (145) POJ (139) Difficult Algorithm (134) EPI (127) LeetCode - Phone (121) Different Solutions (119) Bit Algorithms (116) Lintcode (114) Cracking Coding Interview (110) Smart Algorithm (109) Math (105) HackerRank (89) Binary Search (79) Graph Algorithm (74) Greedy Algorithm (69) Binary Tree (66) DFS (62) Interview Corner (61) List (58) LeetCode - Extended (57) Advanced Data Structure (55) Codility (54) Algorithm Interview (53) BFS (53) ComProGuide (52) Geometry Algorithm (48) USACO (46) Trie (44) Stack (43) Binary Search Tree (42) Mathematical Algorithm (42) ACM-ICPC (41) Data Structure (40) Interval (40) Jobdu (39) Knapsack (39) Recursive Algorithm (38) Space Optimization (38) String Algorithm (38) Matrix (37) Codeforces (36) Introduction to Algorithms (36) Must Known (36) Backtracking (35) Beauty of Programming (35) Sort (35) Union-Find (34) Array (33) prismoskills (33) Segment Tree (32) Sliding Window (32) Data Structure Design (31) HDU (31) Google Code Jam (30) Permutation (30) Puzzles (30) Array O(N) (29) Company-Airbnb (29) Company-Zenefits (28) Microsoft 100 - July (28) to-do-must (28) Palindrome (27) Random (27) Graph (26) GeeksQuiz (25) Logic Thinking (25) hihocoder (25) Company-Facebook (23) High Frequency (23) Pre-Sort (23) Queue (23) TopCoder (23) Algorithm Game (22) Company - LinkedIn (22) Hash (22) Lintcode - Review (22) Priority Queue (22) Binary Indexed Trees (21) DFS + Review (21) Brain Teaser (20) CareerCup (20) Company - Twitter (20) Merge Sort (20) Bisection Method (19) Follow Up (19) Post-Order Traverse (19) UVA (19) LeetCode Hard (18) Probabilities (18) Company-Uber (17) Topological Sort (17) Codercareer (16) Game Theory (16) Heap (16) Ordered Stack (16) Shortest Path (16) String Search (16) Tree Traversal (16) itint5 (16) Difficult (15) Iterator (15) O(N) (15) Two Pointers (15) Binary Search - Bisection (14) Number (14) Number Theory (14) Time Complexity (14) Amazon Interview (13) BST (13) Basic Algorithm (13) Codechef (13) Euclidean GCD (13) KMP (13) Long Increasing Sequence(LIS) (13) Majority (13) mitbbs (13) Combination (12) Computational Geometry (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) Miscs (11) Princeton (11) Reverse Thinking (11) Tree DP (11) X Sum (11) 挑战程序设计竞赛 (11) Coin Change (10) Company - Microsoft (10) Facebook Hacker Cup (10) HackerRank Easy (10) Rolling Hash (10) SPOJ (10) Theory (10) Tutorialhorizon (10) Bucket Sort (9) DFS+Cache (9) DP-Space Optimization (9) Divide and Conquer (9) Kadane - Extended (9) Mathblog (9) Max-Min Flow (9) O(1) Space (9) Prefix Sum (9) Quick Sort (9) Stack Overflow (9) Stock (9) System Design (9) Use XOR (9) Book Notes (8) Bottom-Up (8) Company-Amazon (8) DFS+BFS (8) DP-Multiple Relation (8) LeetCode - DP (8) Longest Common Subsequence(LCS) (8) Prime (8) Suffix Tree (8) Tech-Queries (8) 穷竭搜索 (8) Algorithm Problem List (7) Facebook Interview (7) Fibonacci Numbers (7) Game Nim (7) Graph BFS (7) HackerRank Difficult (7) Hackerearth (7) Interval Tree (7) Inversion (7) Kadane’s Algorithm (7) Level Order Traversal (7) Linked List (7) Math-Divisible (7) Probability DP (7) Quick Select (7) Radix Sort (7) Simulation (7) TreeMap (7) Xpost (7) n00tc0d3r (7) 蓝桥杯 (7) Catalan Number (6) Classic Data Structure Impl (6) DFS+DP (6) DP - Tree (6) DP-Print Solution (6) Dutch Flag (6) How To (6) Interviewstreet (6) Knapsack - MultiplePack (6) Manacher (6) Minimum Spanning Tree (6) Multiple Data Structures (6) One Pass (6) Programming Pearls (6) Rabin-Karp (6) Randomized Algorithms (6) Sampling (6) Schedule (6) Suffix Array (6) Threaded (6) reddit (6) AI (5) Algorithm - Brain Teaser (5) Art Of Programming-July (5) Big Data (5) Brute Force (5) Code Kata (5) Codility-lessons (5) Coding (5) Company - WMware (5) Convex Hull (5) Crazyforcode (5) DP-Include vs Exclude (5) Fast Slow Pointers (5) Graph Cycle (5) Hash Strategy (5) Immutability (5) Java (5) Left and Right Array (5) Matrix Chain Multiplication (5) Maze (5) Microsoft Interview (5) Morris Traversal (5) Pre-Sum (5) Pruning (5) Quadtrees (5) Quick Partition (5) Quora (5) SPFA(Shortest Path Faster Algorithm) (5) Subarray Sum (5) Sweep Line (5) Traversal Once (5) TreeSet (5) Word Search (5) jiuzhang (5) 单调栈 (5) 树形DP (5) 1point3acres (4) Abbreviation (4) Anagram (4) Approximate Algorithm (4) Backtracking-Include vs Exclude (4) Brute Force - Enumeration (4) Chess Game (4) Consistent Hash (4) Cycle (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) LeetCode - Recursive (4) LeetCode - TODO (4) MST (4) MinMax (4) N Queens (4) Nerd Paradise (4) Parallel Algorithm (4) Practical Algorithm (4) Probability (4) Programcreek (4) Spell Checker (4) Stock Maximize (4) Stream (4) Subset Sum (4) Subsets (4) Sudoku (4) Symbol Table (4) Triangle (4) Water Jug (4) algnotes (4) fgdsb (4) to-do-2 (4) 最大化最小值 (4) A Star (3) Algorithm - How To (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) Graph - Bipartite (3) Include vs Exclude (3) Joseph (3) Jump Game (3) K (3) Knapsack-多重背包 (3) LeetCode - Bit (3) Linked List Merge Sort (3) LogN (3) Master Theorem (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) Sieve of Eratosthenes (3) Stack - Smart (3) State Machine (3) Streaming Algorithm (3) Subtree (3) Transform Tree (3) Trie + DFS (3) Two Pointers Window (3) Warshall Floyd (3) With Random Pointer (3) Word Ladder (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 Counting (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 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) Linked Interview (2) Linked List Sort (2) Longest SubArray (2) Lucene-Solr (2) Math-Remainder Queue (2) Matrix Power (2) Median (2) Minimum Vertex Cover (2) Negative All Values (2) Number Each Digit (2) Numerical Method (2) Object Design (2) Order Statistic Tree (2) Parent-Only Tree (2) Parentheses (2) Parser (2) Peak (2) Programming (2) Proof (2) Range Minimum Query (2) Reuse Forward Backward (2) Robot (2) Rosettacode (2) Scan from right (2) Search (2) Shuffle (2) SimHash (2) Simple Algorithm (2) Skyline (2) Spatial Index (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) Yahoo Interview (2) Young Tableau (2) 剑指Offer (2) 数位DP (2) 1-X (1) 51Nod (1) Akka (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 String (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) Cloest (1) Clone (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) Cont Improvement (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) Diagonal (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) Kth Element (1) Least Common Ancestor (1) LeetCode - Binary Tree (1) LeetCode - Coding (1) LeetCode - Detail (1) LeetCode - Related (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) Machine Learning (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) Memory-Efficient (1) MinHash (1) MinMax Heap (1) Monotone Queue (1) Monto Carlo (1) Multi-End BFS (1) Multi-Reverse (1) Multiple DFS (1) Multiple Tasks (1) Next Successor (1) Offline Algorithm (1) PAT (1) Parenthesis (1) Partition (1) Path Finding (1) Patience Sort (1) Persistent (1) Pigeon Hole Principle (1) Power Set (1) Pratical Algorithm (1) PreProcess (1) Probabilistic Data Structure (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) Square (1) Stac (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 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) Virtual Matrix (1) Wiggle Sort (1) Wikipedia (1) ZOJ (1) ZigZag (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