Sunday, March 27, 2016

LeetCode - 220 Contains Duplicate III


http://www.programcreek.com/2014/06/leetcode-contains-duplicate-iii-java/
Given an array of integers, find out whether there are two distinct indices i and j in the array such that the difference between nums[i] and nums[j] is at most t and the difference between i and j is at most k.
X. Bucket Sort: O(n)
To determine whether there is a nearby almost duplicate in one pass we need to keep a sliding window with length kand a dynamic set can do this. Problem is we need to check almost duplicate each time we add a item into the window/set. If we use BST as our dynamic set, the cost of find almost duplicate is O(log n) therefore the whole solution is O(n log n).

Now can we check it in O(1) time on average? The answer is yes. Hash table can do insert, find and remove in O(1) time on average and by carefully design the hash table we can check almost duplicate inO(1) too.
The idea is like the bucket sort algorithm. Suppose we have consecutive buckets covering the range of nums with each bucket a width of t+1. If there are two item with difference <= t, one of the two will happen:
  1. the two in the same bucket
  2. the two in neighbor buckets
Note that we do not need to actually allocate a lot of buckets. At any time there will only be at most min(k, n)buckets. All we need to do is calculate the label of the bucket m = value/(t+1), and check the buckets m - 1m,m + 1. The whole algorithm is then O(n).
For Java, it has a bit of problem because in java -3 / 5 = 0 unlike in python -3 / 5 = -1. We can use a functiongetID to work around this. Or we can get the minimum of nums first and use that as the “zero point”. Or we can simply use Integer.MIN_VALUE as the “zero point”. The two alternatives may need Long in case of overflow.
private long getID(long i, long w) { return i < 0 ? (i + 1) / w - 1 : i / w; } public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) { if (t < 0) return false; Map<Long, Long> d = new HashMap<>(); long w = (long)t + 1; for (int i = 0; i < nums.length; ++i) { long m = getID(nums[i], w); if (d.containsKey(m)) return true; if (d.containsKey(m - 1) && Math.abs(nums[i] - d.get(m - 1)) < w) return true; if (d.containsKey(m + 1) && Math.abs(nums[i] - d.get(m + 1)) < w) return true; d.put(m, (long)nums[i]); if (i >= k) d.remove(getID(nums[i - k], w)); } return false; }

def containsNearbyAlmostDuplicate(self, nums, k, t): if t < 0: return False n = len(nums) d = {} w = t + 1 for i in xrange(n): m = nums[i] / w if m in d: return True if m - 1 in d and abs(nums[i] - d[m - 1]) < w: return True if m + 1 in d and abs(nums[i] - d[m + 1]) < w: return True d[m] = nums[i] if i >= k: del d[nums[i - k] / w] return False

https://leetcode.com/discuss/38206/ac-o-n-solution-in-java-using-buckets-with-explanation
As a followup question, it naturally also requires maintaining a window of size k. When t == 0, it reduces to the previous question so we just reuse the solution.
Since there is now a constraint on the range of the values of the elements to be considered duplicates, it reminds us of doing a range check which is implemented in tree data structure and would take O(LogN) if a balanced tree structure is used, or doing a bucket check which is constant time. We shall just discuss the idea using bucket here.
Bucketing means we map a range of values to the a bucket. For example, if the bucket size is 3, we consider 0, 1, 2 all map to the same bucket. However, if t == 3, (0, 3) is a considered duplicates but does not map to the same bucket. This is fine since we are checking the buckets immediately before and after as well. So, as a rule of thumb, just make sure the size of the bucket is reasonable such that elements having the same bucket is immediately considered duplicates or duplicates must lie within adjacent buckets. So this actually gives us a range of possible bucket size, i.e. t and t + 1. We just choose it to be t and a bucket mapping to be num / t.
Another complication is that negative ints are allowed. A simple num / t just shrinks everything towards 0. Therefore, we can just reposition every element to start from Integer.MIN_VALUE.

    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        if (k < 1 || t < 0) return false;
        Map<Long, Long> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            long remappedNum = (long) nums[i] - Integer.MIN_VALUE;
            long bucket = remappedNum / ((long) t + 1);
            if (map.containsKey(bucket)
                    || (map.containsKey(bucket - 1) && remappedNum - map.get(bucket - 1) <= t)
                        || (map.containsKey(bucket + 1) && map.get(bucket + 1) - remappedNum <= t))
                            return true;
            if (map.entrySet().size() >= k) {
                long lastBucket = ((long) nums[i - k] - Integer.MIN_VALUE) / ((long) t + 1);
                map.remove(lastBucket);
            }
            map.put(bucket, remappedNum);
        }
        return false;
    }
May I ask why it's necessary to reposition every element to start from Integer.MIN_VALUE?
I suppose it is not necessary if you use a different way to do the assignment of buckets. In this case, the elements are assigned to a bucket by a simple division of a bucket size. Hence, it will, for example, assign both -2 and +2 to bucket 0 if the bucket size is > 2. If we want that elements having the same bucket 0 are immediately considered duplicates, such assignment would be undesirable. Hence, a repositioning is used to cope with this. And starting from MIN_VALUE is just because of the constraints on the input range.

X. Using TreeSet O(NlogK)
- the treeSet's size is k


public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
 if (k < 1 || t < 0)
  return false;
 
 TreeSet<Integer> set = new TreeSet<Integer>();
 
 for (int i = 0; i < nums.length; i++) {
  int c = nums[i];
  if ((set.floor(c) != null && c <= set.floor(c) + t)
  || (set.ceiling(c) != null && c >= set.ceiling(c) -t))
   return true;
 
  set.add(c); // remove first then add => corner case: nums[i - k]=c
 
  if (i >= k)
   set.remove(nums[i - k]);
 }
 
 return false;
}
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
 if (k < 1 || t < 0)
  return false;
 
 SortedSet<Long> set = new TreeSet<Long>();
 
 for (int j = 0; j < nums.length; j++) {
  long leftBoundary = (long) nums[j] - t;
  long rightBoundary = (long) nums[j] + t + 1;
  SortedSet<Long> subSet = set.subSet(leftBoundary, rightBoundary);
 
  if (!subSet.isEmpty())
   return true;
 
  set.add((long) nums[j]);
 
  if (j >= k) {
   set.remove((long) nums[j - k]);
  }
 }
 
 return false;
}
https://leetcode.com/discuss/68090/easy-ac-solution-using-treeset-long-in-java
Use subset
The time complexity of TreeSet.subSet is O(1) because it essentially creates a wrapper around the original set. However, its isEmpty operation is not instant because it searches for successors/predecessors, which, I believe, is O(log k). So the complexity is O(n log k) like other tree-based solutions.
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) { if (nums == null || nums.length == 0) return false; TreeSet<Long> set = new TreeSet<>(); set.add((long) nums[0]); for (int i = 1; i < nums.length; i++) { if (i > k) set.remove((long) nums[i - k - 1]);//\\ long left = (long) nums[i] - t; long right = (long) nums[i] + t; if (left <= right && !set.subSet(left, right + 1).isEmpty()) return true; set.add((long) nums[i]); } return false; }
https://discuss.leetcode.com/topic/15191/java-o-n-lg-k-solution
This problem requires to maintain a window of size k of the previous values that can be queried for value ranges. The best data structure to do that is Binary Search Tree. As a result maintaining the tree of size k will result in time complexity O(N lg K). In order to check if there exists any value of range abs(nums[i] - nums[j]) to simple queries can be executed both of time complexity O(lg K)
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        if (nums == null || nums.length == 0 || k <= 0) {
            return false;
        }

        final TreeSet<Integer> values = new TreeSet<>();
        for (int ind = 0; ind < nums.length; ind++) {

            final Integer floor = values.floor(nums[ind] + t);
            final Integer ceil = values.ceiling(nums[ind] - t);
            if ((floor != null && floor >= nums[ind])
                    || (ceil != null && ceil <= nums[ind])) {
                return true;
            }

            values.add(nums[ind]);
            if (ind >= k) {
                values.remove(nums[ind - k]);
            }
        }

        return false;
    }
TODO:
how about save one ceiling operation

because if floor <= nums[ind] + t and floor >= nums[ind] - t, then that's a duplicate. If floor < nums[ind] - t, there won't be a duplicate because floor is the biggest value in the treeset that is less than (nums[ind] + t). Hope it helps.
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) { if (nums == null || nums.length == 0 || k <= 0) { return false; } final TreeSet<Long> values = new TreeSet<>(); for (int ind = 0; ind < nums.length; ind++) { Long floor = values.floor((long)nums[ind] + t); if (floor != null && floor + t >= nums[ind]) { return true; } values.add((long)nums[ind]); if (ind >= k) { values.remove((long)nums[ind - k]); } } return false; }

X. Different
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) { if(nums.length<2||k<1||t<0) return false; ValuePosPair[] valPosArr = new ValuePosPair[nums.length]; for(int i =0;i<nums.length;i++) valPosArr[i] = new ValuePosPair(nums[i],i); Arrays.sort(valPosArr); for(int i=0;i<valPosArr.length;i++){ for(int j=i+1;j<valPosArr.length&&((long)valPosArr[j].val-(long)valPosArr[i].val<=(long)t);j++){ if(Math.abs(valPosArr[j].pos-valPosArr[i].pos)<=k) return true; } } return false; }
class ValuePosPair implements Comparable<ValuePosPair>{ int val; int pos; ValuePosPair(int v, int p) { val = v; pos = p;} public int compareTo(ValuePosPair x){ return this.val - x.val; } }


No comments:

Post a Comment

Labels

GeeksforGeeks (976) Algorithm (811) LeetCode (654) to-do (599) Review (362) Classic Algorithm (334) Classic Interview (298) Dynamic Programming (263) Google Interview (233) LeetCode - Review (233) Tree (146) POJ (137) Difficult Algorithm (136) EPI (127) Different Solutions (119) 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) Binary Tree (58) List (58) 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) Jobdu (39) Interval (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) Space Optimization (34) Array (33) Trie (33) prismoskills (33) Backtracking (32) Segment Tree (32) Union-Find (32) 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 (27) GeeksQuiz (25) Logic Thinking (25) hihocoder (25) High Frequency (23) Palindrome (23) Algorithm Game (22) Company - LinkedIn (22) Graph (22) Hash (22) Queue (22) DFS + Review (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) Bisection Method (14) Difficult (14) Number (14) Number Theory (14) Post-Order Traverse (14) Priority Quieue (14) Amazon Interview (13) BST (13) Basic Algorithm (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) Divide and Conquer (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) Graph BFS (8) LeetCode - DP (8) LeetCode Hard (8) Prefix Sum (8) Prime (8) Suffix Tree (8) System Design (8) Tech-Queries (8) Time Complexity (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) 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) Kadane’s Algorithm (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) 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) 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) Sweep Line (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) 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) Master Theorem (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) 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) Parent-Only Tree (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