Monday, March 20, 2017

LeetCode 532 - K-diff Pairs in an Array


https://leetcode.com/problems/k-diff-pairs-in-an-array/
Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.
Example 1:
Input: [3, 1, 4, 1, 5], k = 2
Output: 2
Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of unique pairs.
Example 2:
Input:[1, 2, 3, 4, 5], k = 1
Output: 4
Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).
Example 3:
Input: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).
Note:
  1. The pairs (i, j) and (j, i) count as the same pair.
  2. The length of the array won't exceed 10,000.
  3. All the integers in the given input belong to the range: [-1e7, 1e7].
X.
https://wlypku.github.io/2017/03/05/Leetcode-week22/
* for every number in the array:
* - if there was a number previously k-diff with it, save the smaller to a set;
* - and save the value to a set;
*/
int findPairs(vector<int>& nums, int k) {
if (k < 0) {
return 0;
}
set<int> starters;
set<int> indices;
for (int i = 0; i < nums.size(); i++) {
if (indices.find(nums[i] - k) != indices.end()) {
starters.insert(nums[i] - k);
}
if (indices.find(nums[i] + k) != indices.end()) {
starters.insert(nums[i]);
}
indices.insert(nums[i]);
}
return starters.size();
}
X.
https://discuss.leetcode.com/topic/81714/java-o-n-solution-one-hashmap-easy-to-understand
将数组值用hashmap来存起来,value来存储个数。如果k=0的话,value>=2的时候count+1,如果k不等于0,则看key+k的值在map里面是否包含,包含的话就count+1
    public int findPairs(int[] nums, int k) {
        if (nums == null || nums.length == 0 || k < 0)   return 0;
        
        Map<Integer, Integer> map = new HashMap<>();
        int count = 0;
        for (int i : nums) {
            map.put(i, map.getOrDefault(i, 0) + 1);
        }
        
        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            if (k == 0) {
                //count how many elements in the array that appear more than twice.
                if (entry.getValue() >= 2) {
                    count++;
                } 
            } else {
                if (map.containsKey(entry.getKey() + k)) {
                    count++;
                }
            }
        }
        
        return count;
    }
https://discuss.leetcode.com/topic/81714/java-o-n-solution-one-hashmap-easy-to-understand/7
    public int findPairs(int[] nums, int k) {
        Map<Integer,Integer> mp = new HashMap<Integer,Integer>();
        if(k<0)
            return 0;
            
        for(int i=0;i<nums.length;i++) {
            mp.put(nums[i],i);
        }
        
        int cnt=0;
        for(int i=0;i<nums.length;i++) {
            if(mp.containsKey(k+nums[i]) && mp.get(k+nums[i])!=i) {
                cnt++;
                mp.remove(k+nums[i]);
            }
        }
        return cnt;
    }
X.
https://discuss.leetcode.com/topic/81745/two-pointer-approach
   public int findPairs(int[] nums, int k) {
        Arrays.sort(nums);

        int start = 0, end = 1, result = 0;
        while (start < nums.length && end < nums.length) {
            if (start == end || nums[start] + k > nums[end]) {
                end++;
            } else if (nums[start] + k < nums[end]) {
                start++;
            } else {
                start++;
                result++;
                // start
                //  |
                // [1, 1, ...., 8, 8]
                //              |
                //             end
                while (start < nums.length && nums[start] == nums[start - 1]) start++;
                end = Math.max(end + 1, start + 1);
            }
        }
        return result;
    }
The problem is just a variant of 2-sum.
Update: Fixed a bug that can cause integer subtraction overflow.
Update: The code runs in O(n log n) time, using O(1) space.
public int findPairs(int[] nums, int k) {
    int ans = 0;
    Arrays.sort(nums);
    for (int i = 0, j = 0; i < nums.length; i++) {
        for (j = Math.max(j, i + 1); j < nums.length && (long) nums[j] - nums[i] < k; j++) ;
        if (j < nums.length && (long) nums[j] - nums[i] == k) ans++;
        while (i + 1 < nums.length && nums[i] == nums[i + 1]) i++;
    }
    return ans;
}
https://discuss.leetcode.com/topic/81678/self-explained-ac-java-sliding-window
 public  int findPairs(int[] nums, int k) {
 if(k<0 || nums.length<=1){
     return 0;
 }
   
         Arrays.sort(nums);
         int count = 0;
         int left = 0;
         int right = 1;
         
         while(right<nums.length){
             int firNum = nums[left];
             int secNum = nums[right];
             // If less than k, increase the right index
             if(secNum-firNum<k){
                 right++;
             }
             // If larger than k, increase the left index
             else if(secNum - firNum>k){
                 left++;   
             }
             // If equal, move left and right to next different number
             else{
                 count++;
                 while(left<nums.length && nums[left]==firNum){
                     left++;
                 }
                 while(right<nums.length && nums[right]==secNum){
                     right++;
                 }
                             
             }
             //left and right should not be the same number
             if(right==left){
              right++;
             }
         }
        return count;
    }
https://discuss.leetcode.com/topic/81897/simple-idea-o-nlogn-time-o-1-space-java-solution
    public int findPairs(int[] nums, int k) {
        if (nums == null || nums.length < 2) return 0;
        int res = 0;
        Arrays.sort(nums);
        for (int i = 0; i < nums.length; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            if (helper(nums, i + 1, nums[i] + k)) res++;
        }
        return res;
    }
    private boolean helper(int[] nums, int l, int target) {
        int r = nums.length - 1;
        while (l <= r) {
            int mid = l + (r - l) / 2;
            if (nums[mid] == target) return true;
            if (nums[mid] < target) l = mid + 1;
            else r = mid - 1;
        }
        return false;
    }
X. Using hashset
https://discuss.leetcode.com/topic/81690/short-java-solution-but-two-hashsets
    public int findPairs(int[] nums, int k) {
        Arrays.sort(nums);
        Set<Integer> seenNum = new HashSet<>();
        Set<String> seenPair = new HashSet<>();
        int result = 0;
        
        for (int i = 0; i < nums.length; i++) {
            int prev = nums[i] - k;
            if (seenNum.contains(prev) && !seenPair.contains(prev + "," + nums[i])) {
                result++;
                seenPair.add(prev + "," + nums[i]);
            }
            seenNum.add(nums[i]);
        }
        return result;
    }
http://blog.csdn.net/zhouziyu2011/article/details/60466898
成对的值不分先后,所以先对nums进行排序。
用一个set存储出现过的值,用于后续判断是否某个值已经有值与其成对。
分为两种情况:
(1)k==0,即找出值相等的对数。
再用一个sameSet存储所有已成对的值,避免同一个值加入结果多次。只有sameSet中不含该值,且set中包含了该值,才能加入结果。
(2)k!=0,即找出差的绝对值为k的对数。
只有set中不包含该值但包含了该值-k,才能加入结果。

  1.     public int findPairs(int[] nums, int k) {  
  2.         int len = nums.length, result = 0;  
  3.         Arrays.sort(nums);  
  4.         Set<Integer> set = new HashSet<Integer>();  
  5.         Set<Integer> sameSet = new HashSet<Integer>();  
  6.         if (k != 0) {  
  7.             for (int i = 0; i < len; i++) {  
  8.                 if (!set.contains(nums[i]) && set.contains(nums[i] - k))   
  9.                     result++;  
  10.                 set.add(nums[i]);  
  11.             }  
  12.         }  
  13.         else {  
  14.             for (int i = 0; i < len; i++) {  
  15.                 if (!sameSet.contains(nums[i]) && set.contains(nums[i])) {  
  16.                     result++;  
  17.                     sameSet.add(nums[i]);  
  18.                 }  
  19.                 set.add(nums[i]);  
  20.             }   
  21.         }  
  22.         return result;  
  23.     } 
http://www.techiedelight.com/find-pairs-with-given-difference-array/
Given an unsorted array of integers, print all pairs with given difference k in it.
arr = [1, 5, 2, 2, 2, 5, 5, 4]
k = 3

Output:
(2, 5) and (1, 4)

void findPair(int arr[], int n, int diff)
{
    // array is unsorted
    // take an empty set
    unordered_set<int> set;
    // do for each element in the array
    for (int i = 0; i < n; i++)
    {
        // check if pair with given difference (arr[i], arr[i]-diff) exists
        if (set.find(arr[i] - diff) != set.end())
            cout << "(" << arr[i] << ", " << arr[i] - diff<< ")\n";
        // check if pair with given difference (arr[i]+diff, arr[i]) exists
        if (set.find(arr[i] + diff) != set.end())
            cout << "(" << arr[i] + diff << ", " << arr[i] << ")\n";
        // insert element into the set
        set.insert(arr[i]);
    }
}

We can handle duplicates pairs by sorting the array first and then skipping adjacent similar elements.
void findPair(int arr[], int n, int diff)
{
    // sort array in ascending order
    sort(arr, arr + n);
    // take an empty set
    unordered_set<int> set;
    // do for each element in the array
    for (int i = 0; i < n; i++)
    {
        // to avoid printing duplicates (skip adjacent duplicates)
        while (i < n && arr[i] == arr[i+1])
            i++;
        // check if pair with given difference (arr[i], arr[i]-diff) exists
        if (set.find(arr[i] - diff) != set.end())
            cout << "(" << arr[i] << ", " << arr[i] - diff << ")\n";
        // check if pair with given difference (arr[i]+diff, arr[i]) exists
        if (set.find(arr[i] + diff) != set.end())
            cout << "(" << arr[i] + diff << ", " << arr[i] << ")\n";
        // insert element into the set
        set.insert(arr[i]);
    }
}

We can avoid using extra space by performing binary search for element (arr[i] – diff) or (arr[i] + diff) instead of using hashing.
void findPair(int arr[], int n, int diff)
{
    // sort array in ascending order
    sort(arr, arr + n);
    // do for each element in the array
    for (int i = 0; i < n; i++)
    {
        // to avoid printing duplicates (skip adjacent duplicates)
        while (i < n && arr[i] == arr[i+1])
            i++;
        // perform binary search for element (arr[i] - diff)
        if (binary_search(arr, arr + n, arr[i] - diff))    // STL binary search
            cout << "(" << arr[i] << ", " << arr[i] - diff << ")\n";
    }
}

void findPair(int arr[], int n, int diff)
{
    // sort array in ascending order
    sort(arr, arr + n);
    // maintain two indexes in the array
    int i = 0, j = 0;
    // run till end of array is reached
    while (i < n && j < n)
    {
        // to avoid printing duplicates
        while (i < n && arr[i] == arr[i+1])
            i++;
        while (j < n && arr[j] == arr[j+1])
            j++;
        // increment i if current difference is more than the desired difference
        if (arr[j] - arr[i] > diff)
            i++;
        // increment j if current difference is less than the desired difference
        else if (arr[j] - arr[i] < diff)
            j++;
        // print the pair and increment both i, j if current difference is same
        // as the desired difference
        else
        {
            cout << "(" << arr[j] << ", " << arr[i] << ")\n";
            i++, j++;
        }
    }
}

No comments:

Post a Comment

Labels

GeeksforGeeks (1109) LeetCode (1095) Review (846) Algorithm (795) to-do (633) LeetCode - Review (574) Classic Algorithm (324) Dynamic Programming (294) Classic Interview (288) Google Interview (242) Tree (145) POJ (139) Difficult Algorithm (132) LeetCode - Phone (127) EPI (125) Bit Algorithms (120) Different Solutions (120) Lintcode (113) Cracking Coding Interview (110) Smart Algorithm (109) Math (108) HackerRank (89) Binary Search (83) Binary Tree (82) Graph Algorithm (76) Greedy Algorithm (73) DFS (71) Stack (65) LeetCode - Extended (62) Interview Corner (61) List (58) Advanced Data Structure (56) BFS (54) Codility (54) ComProGuide (52) Algorithm Interview (50) Geometry Algorithm (49) Trie (49) Binary Search Tree (47) USACO (46) Interval (45) LeetCode Hard (42) Mathematical Algorithm (42) ACM-ICPC (41) Data Structure (40) Knapsack (40) Space Optimization (40) Jobdu (39) Matrix (39) Recursive Algorithm (39) String Algorithm (38) Union-Find (37) Backtracking (36) Codeforces (36) Introduction to Algorithms (36) Must Known (36) Beauty of Programming (35) Sort (35) Array (33) Data Structure Design (33) Segment Tree (33) Sliding Window (33) prismoskills (33) HDU (31) Priority Queue (31) Google Code Jam (30) Permutation (30) Puzzles (30) Array O(N) (29) Company-Airbnb (29) Company-Zenefits (28) Microsoft 100 - July (28) Palindrome (28) to-do-must (28) Graph (27) Random (27) Company - LinkedIn (25) GeeksQuiz (25) Logic Thinking (25) Post-Order Traverse (25) Pre-Sort (25) Time Complexity (25) hihocoder (25) Queue (24) Company-Facebook (23) High Frequency (23) TopCoder (23) Algorithm Game (22) Binary Indexed Trees (22) Bisection Method (22) Hash (22) DFS + Review (21) Lintcode - Review (21) Two Pointers (21) Brain Teaser (20) CareerCup (20) Company - Twitter (20) LeetCode - DP (20) Merge Sort (20) O(N) (20) Follow Up (19) UVA (19) Ordered Stack (18) Probabilities (18) Company-Uber (17) Game Theory (17) Topological Sort (17) Codercareer (16) Heap (16) Shortest Path (16) String Search (16) Tree Traversal (16) itint5 (16) Difficult (15) Iterator (15) BST (14) KMP (14) LeetCode - DFS (14) Number (14) Number Theory (14) Amazon Interview (13) Basic Algorithm (13) Codechef (13) Euclidean GCD (13) Long Increasing Sequence(LIS) (13) Majority (13) Reverse Thinking (13) mitbbs (13) Combination (12) Computational Geometry (12) LeetCode - Classic (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) Miscs (11) Princeton (11) Proof (11) Rolling Hash (11) Tree DP (11) X Sum (11) 挑战程序设计竞赛 (11) Bisection (10) Bucket Sort (10) Coin Change (10) Company - Microsoft (10) DFS+Cache (10) Facebook Hacker Cup (10) HackerRank Easy (10) O(1) Space (10) SPOJ (10) Theory (10) TreeMap (10) Tutorialhorizon (10) DP-Multiple Relation (9) DP-Space Optimization (9) Divide and Conquer (9) Kadane - Extended (9) Mathblog (9) Max-Min Flow (9) Prefix Sum (9) Quick Sort (9) Simulation (9) Stack Overflow (9) Stock (9) System Design (9) Use XOR (9) Book Notes (8) Bottom-Up (8) Company-Amazon (8) DFS+BFS (8) Interval Tree (8) Left and Right Array (8) Linked List (8) Longest Common Subsequence(LCS) (8) Prime (8) Suffix Tree (8) Tech-Queries (8) Traversal Once (8) 穷竭搜索 (8) Algorithm Problem List (7) Expression (7) Facebook Interview (7) Fibonacci Numbers (7) Game Nim (7) Graph BFS (7) HackerRank Difficult (7) Hackerearth (7) Inversion (7) Kadane’s Algorithm (7) Level Order Traversal (7) Math-Divisible (7) Probability DP (7) Quick Select (7) Radix Sort (7) TreeSet (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) Dijkstra (6) Dutch Flag (6) How To (6) Interviewstreet (6) Knapsack - MultiplePack (6) Manacher (6) Minimum Spanning Tree (6) Morris Traversal (6) Multiple Data Structures (6) One Pass (6) Programming Pearls (6) Pruning (6) Rabin-Karp (6) Randomized Algorithms (6) Sampling (6) Schedule (6) Stream (6) Suffix Array (6) Threaded (6) Xpost (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) Cycle (5) DP-Include vs Exclude (5) Fast Slow Pointers (5) Find Rule (5) Graph Cycle (5) Hash Strategy (5) Immutability (5) Java (5) LeetCode - TODO (5) Matrix Chain Multiplication (5) Maze (5) Microsoft Interview (5) Pre-Sum (5) Quadtrees (5) Quick Partition (5) Quora (5) SPFA(Shortest Path Faster Algorithm) (5) Subarray Sum (5) Sudoku (5) Sweep Line (5) Word Search (5) jiuzhang (5) 单调栈 (5) 树形DP (5) 1point3acres (4) Abbreviation (4) Anagram (4) Anagrams (4) Approximate Algorithm (4) Backtracking-Include vs Exclude (4) Brute Force - Enumeration (4) Chess Game (4) Consistent Hash (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) 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) Subset Sum (4) Subsets (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) B Tree (3) Big Data Algorithm (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) Finite Automata (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) O(N) Hard (3) Online Algorithm (3) Parser (3) Pascal's Triangle (3) Pattern Match (3) Project Euler (3) Rectangle (3) Scala (3) SegmentFault (3) Shuffle (3) Sieve of Eratosthenes (3) Skyline (3) Stack - Smart (3) State Machine (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) Binary Search - Smart (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) Company-Snapchat (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) Edit Distance (2) Factor (2) Forward && Backward Scan (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) 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) Peak (2) Programming (2) Range Minimum Query (2) Regular Expression (2) Return Multiple Values (2) Reuse Forward Backward (2) Robot (2) Rosettacode (2) Scan from right (2) Search (2) SimHash (2) Simple Algorithm (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) Tree Without Tree Predefined (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) Augmented BST (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 Sarch Tree (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) Clean Code (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-Yelp (1) Compression Algorithm (1) Concise (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 - Facebook (1) Isomorphic (1) JDK8 (1) K Dimensional Tree (1) Knapsack - Fractional (1) Knapsack - ZeroOnePack (1) Knight (1) Knuth Shuffle (1) Kosaraju’s algorithm (1) Kruskal (1) Kth Element (1) Least Common Ancestor (1) LeetCode - Binary Tree (1) LeetCode - Coding (1) LeetCode - Construction (1) LeetCode - Detail (1) LeetCode - Related (1) LeetCode - Thinking (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) Maintain State (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 Element (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) Resources (1) Reverse Inorder Traversal (1) Robin (1) Selection (1) Self Balancing BST (1) Similarity (1) Sort && Binary Search (1) Square (1) Streaming Algorithm (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) Test Cases (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) Two-End-BFS (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