Friday, September 16, 2016

Min-Max Range Queries - GeeksforGeeks


Min-Max Range Queries - GeeksforGeeks
Given an array arr[0 . . . n-1]. We need to efficiently find the minimum and maximum value from index qs (query start) to qe (query end) where 0 <= qs <= qe <= n-1. We are given multiple queries.

Efficient solution : This problem can be solved more efficiently by using Segment Tree
Can we do better if there are no updates on array?
The above segment tree based solution also allows array updates also to happen in O(Log n) time. Assume a situation when there are no updates (or array is static). We can actually process all queries in O(1) time with some preprocessing. One simple solution is to make a 2D table of nodes that stores all range minimum and maximum. This solution requires O(1) query time, but requires O(n2) preprocessing time and O(n2) extra space which can be a problem for large n. We can solve this problem in O(1) query time, O(n Log n) space and O(n Log n) preprocessing time using Sparse Table.
// Node for storing minimum nd maximum value of given range
struct node
{
   int minimum;
   int maximum;
};
 
// A utility function to get the middle index from corner indexes.
int getMid(int s, int e) {  return s + (e -s)/2;  }
 
/*  A recursive function to get the minimum and maximum value in
     a given range of array indexes. The following are parameters
     for this function.
 
    st    --> Pointer to segment tree
    index --> Index of current node in the segment tree. Initially
              0 is passed as root is always at index 0
    ss & se  --> Starting and ending indexes of the segment
                  represented  by current node, i.e., st[index]
    qs & qe  --> Starting and ending indexes of query range */
struct node MaxMinUntill(struct node *st, int ss, int se, int qs,
                         int qe, int index)
{
    // If segment of this node is a part of given range, then return
    //  the minimum and maximum node of the segment
    struct node tmp,left,right;
    if (qs <= ss && qe >= se)
        return st[index];
 
    // If segment of this node is outside the given range
    if (se < qs || ss > qe)
    {
       tmp.minimum = INT_MAX;
       tmp.maximum = INT_MIN;
       return tmp;
     }
 
    // If a part of this segment overlaps with the given range
    int mid = getMid(ss, se);
    left = MaxMinUntill(st, ss, mid, qs, qe, 2*index+1);
    right = MaxMinUntill(st, mid+1, se, qs, qe, 2*index+2);
    tmp.minimum = min(left.minimum, right.minimum);
    tmp.maximum = max(left.maximum, right.maximum);
    return tmp;
}
 
// Return minimum and maximum of elements in range from index
// qs (quey start) to qe (query end).  It mainly uses
// MaxMinUtill()
struct node MaxMin(struct node *st, int n, int qs, int qe)
{
    struct node tmp;
 
    // Check for erroneous input values
    if (qs < 0 || qe > n-1 || qs > qe)
    {
        printf("Invalid Input");
        tmp.minimum = INT_MIN;
        tmp.minimum = INT_MAX;
        return tmp;
    }
 
    return MaxMinUntill(st, 0, n-1, qs, qe, 0);
}
 
// A recursive function that constructs Segment Tree for array[ss..se].
// si is index of current node in segment tree st
void constructSTUtil(int arr[], int ss, int se, struct node *st,
                     int si)
{
    // If there is one element in array, store it in current node of
    // segment tree and return
    if (ss == se)
    {
        st[si].minimum = arr[ss];
        st[si].maximum = arr[ss];
        return ;
    }
 
    // If there are more than one elements, then recur for left and
    // right subtrees and store the minimum and maximum of two values
    // in this node
    int mid = getMid(ss, se);
    constructSTUtil(arr, ss, mid, st, si*2+1);
    constructSTUtil(arr, mid+1, se, st, si*2+2);
 
    st[si].minimum = min(st[si*2+1].minimum, st[si*2+2].minimum);
    st[si].maximum = max(st[si*2+1].maximum, st[si*2+2].maximum);
}
 
/* Function to construct segment tree from given array. This function
   allocates memory for segment tree and calls constructSTUtil() to
   fill the allocated memory */
struct node *constructST(int arr[], int n)
{
    // Allocate memory for segment tree
 
    // Height of segment tree
    int x = (int)(ceil(log2(n)));
 
    // Maximum size of segment tree
    int max_size = 2*(int)pow(2, x) - 1;
 
    struct node *st = new struct node[max_size];
 
    // Fill the allocated memory st
    constructSTUtil(arr, 0, n-1, st, 0);
 
    // Return the constructed segment tree
    return st;
}
We have an array arr[0 . . . n-1]. We should be able to efficiently find the minimum value from index L (query start) to R (query end) where 0 <= L <= R <= n-1. Consider a situation when there are many range queries.

Method 2 (Square Root Decomposition)
We can use Square Root Decompositions to reduce space required in above method.
Preprocessing:
1) Divide the range [0, n-1] into different blocks of √n each.
2) Compute minimum of every block of size √n and store the results.
Preprocessing takes O(√n * √n) = O(n) time and O(√n) space.
Query:
1) To query a range [L, R], we take minimum of all blocks that lie in this range. For left and right corner blocks which may partially overlap with given range, we linearly scan them to find minimum.
rmq3
Time complexity of query is O(√n). Note that we have minimum of middle block directly accessible and there can be at most O(√n) middle blocks. There can be atmost two corner blocks that we may have to scan, so we may have to scan 2*O(√n) elements of corner blocks. Therefore, overall time complexity is O(√n).

Method 3 (Sparse Table Algorithm)
The above solution requires only O(√n) space, but takes O(√n) time to query. Sparse table method supports query time O(1) with extra space O(n Log n).
The idea is to precompute minimum of all subarrays of size 2j where j varies from 0 to Log n. Like method 1, we make a lookup table. Here lookup[i][j] contains minimum of range starting from i and of size 2j. For example lookup[0][3] contains minimum of range [0, 7] (starting with 0 and of size 23)
Preprocessing:
How to fill this lookup table? The idea is simple, fill in bottom up manner using previously computed values.
For example, to find minimum of range [0, 7], we can use minimum of following two.
a) Minimum of range [0, 3]
b) Minimum of range [4, 7]
Based on above example, below is formula,
// If arr[lookup[0][3]] <=  arr[lookup[4][7]], 
// then lookup[0][7] = lookup[0][3]
If arr[lookup[i][j-1]] <= arr[lookup[i+2j-1-1][j-1]]
   lookup[i][j] = lookup[i][j-1]

// If arr[lookup[0][3]] >  arr[lookup[4][7]], 
// then lookup[0][7] = lookup[4][7]
Else 
   lookup[i][j] = lookup[i+2j-1-1][j-1] 
rmqSparseTable
Query: 
For any arbitrary range [l, R], we need to use ranges which are in powers of 2. The idea is to use closest power of 2. We always need to do at most comparison (compare minimum of two ranges which are powers of 2). One range starts with L and and ends with “L + closest-power-of-2″. The other range ends at R and starts with “R – same-closest-power-of-2 + 1″. For example, if given range is (2, 10), we compare minimum of two ranges (2, 9) and (3, 10).
Based on above example, below is formula,
// For (2,10), j = floor(Log2(10-2+1)) = 3
j = floor(Log(R-L+1))

// If arr[lookup[0][7]] <=  arr[lookup[3][10]], 
// then RMQ(2,10) = lookup[0][7]
If arr[lookup[L][j]] <= arr[lookup[R-(int)pow(2,j)+1][j]]
   RMQ(L, R) = lookup[L][j]

// If arr[lookup[0][7]] >  arr[lookup[3][10]], 
// then RMQ(2,10) = lookup[3][10]
Else 
   RMQ(L, R) = lookup[i+2j-1-1][j-1]
Since we do only one comparison, time complexity of query is O(1).
So sparse table method supports query operation in O(1) time with O(n Log n) preprocessing time and O(n Log n) space.
// lookup[i][j] is going to store index of minimum value in
// arr[i..j]. Ideally lookup table size should not be fixed and
// should be determined using n Log n. It is kept constant to
// keep code simple.
int lookup[MAX][MAX];
 
// Structure to represent a query range
struct Query
{
    int L, R;
};
 
// Fills lookup array lookup[][] in bottom up manner.
void preprocess(int arr[], int n)
{
    // Initialize M for the intervals with length 1
    for (int i = 0; i < n; i++)
        lookup[i][0] = i;
 
    // Compute values from smaller to bigger intervals
    for (int j=1; (1<<j)<=n; j++)
    {
        // Compute minimum value for all intervals with size 2^j
        for (int i=0; (i+(1<<j)-1) < n; i++)
        {
            // For arr[2][10], we compare arr[lookup[0][7]] and
            // arr[lookup[3][10]]
            if (arr[lookup[i][j-1]] < arr[lookup[i + (1<<(j-1))][j-1]])
                lookup[i][j] = lookup[i][j-1];
            else
                lookup[i][j] = lookup[i + (1 << (j-1))][j-1];     
        }
    }
}
 
// Returns minimum of arr[L..R]
int query(int arr[], int L, int R)
{
    // For [2,10], j = 3
    int j = (int)log2(R-L+1);
 
    // For [2,10], we compare arr[lookup[0][7]] and
    // arr[lookup[3][10]],
    if (arr[lookup[L][j]] <= arr[lookup[R-(int)pow(2,j)+1][j]])
        return arr[lookup[L][j]];
 
   else return arr[lookup[R-(int)pow(2,j)+1][j]];
}
 
// Prints minimum of given m query ranges in arr[0..n-1]
void RMQ(int arr[], int n, Query q[], int m)
{
    // Fills table lookup[n][Log n]
    preprocess(arr, n);
 
    // One by one compute sum of all queries
    for (int i=0; i<m; i++)
    {
        // Left and right boundaries of current range
        int L = q[i].L, R = q[i].R;
 
        // Print sum of current query range
        cout << "Minimum of [" << L << ", "
             << R << "] is "  << query(arr, L, R) << endl;
    }
}


This approach supports query in O(1), but preprocessing takes O(n2time. Also, this approach needs O(n2) extra space which may become huge for large input arrays.
// lookup[i][j] is going to store index of minimum value in
// arr[i..j]
int lookup[MAX][MAX];
 
// Structure to represent a query range
struct Query
{
    int L, R;
};
 
// Fills lookup array lookup[n][n] for all possible values of
// query ranges
void preprocess(int arr[], int n)
{
    // Initialize lookup[][] for the intervals with length 1
    for (int i = 0; i < n; i++)
        lookup[i][i] = i;
 
    // Fill rest of the entries in bottom up manner
    for (int i=0; i<n; i++)
    {
        for (int j = i+1; j<n; j++)
 
           // To find minimum of [0,4], we compare minimum of
           // arr[lookup[0][3]] with arr[4].
           if (arr[lookup[i][j - 1]] < arr[j])
              lookup[i][j] = lookup[i][j - 1];
           else
              lookup[i][j] = j;
    }
}
 
// Prints minimum of given m query ranges in arr[0..n-1]
void RMQ(int arr[], int n, Query q[], int m)
{
    // Fill lookup table for all possible input queries
    preprocess(arr, n);
 
    // One by one compute sum of all queries
    for (int i=0; i<m; i++)
    {
        // Left and right boundaries of current range
        int L = q[i].L, R = q[i].R;
 
        // Print sum of current query range
        cout << "Minimum of [" << L << ", "
             << R << "] is "  << arr[lookup[L][R]] << endl;
    }
}
Simple Solution : We solve this problem using Tournament Method for each query. Complexity for this approach will be O(queries * n).
http://www.geeksforgeeks.org/maximum-and-minimum-in-an-array/
Maximum and minimum of an array using minimum number of comparisons

Write a C function to return minimum and maximum in an array. You program should make minimum number of comparisons.
METHOD 2 (Tournament Method)
Divide the array into two parts and compare the maximums and minimums of the the two parts to get the maximum and the minimum of the the whole array.
Pair MaxMin(array, array_size)
   if array_size = 1
      return element as both max and min
   else if arry_size = 2
      one comparison to determine max and min
      return that pair
   else    /* array_size  > 2 */
      recur for max and min of left half
      recur for max and min of right half
      one comparison determines true max of the two candidates
      one comparison determines true min of the two candidates
      return the pair of max and min



No comments:

Post a Comment

Labels

GeeksforGeeks (959) Algorithm (811) LeetCode (635) to-do (597) Review (337) Classic Algorithm (334) Classic Interview (299) Dynamic Programming (263) Google Interview (232) LeetCode - Review (227) 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) Greedy Algorithm (61) Interview Corner (61) List (58) Binary Tree (56) DFS (56) Algorithm Interview (53) Advanced Data Structure (52) Codility (52) ComProGuide (52) LeetCode - Extended (47) USACO (46) Geometry Algorithm (45) 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