Sunday, May 22, 2016

LeetCode 350 - Intersection of Two Arrays II


https://leetcode.com/problems/intersection-of-two-arrays-ii/
Given two arrays, write a function to compute their intersection.
Example:
Given nums1 = [1, 2, 2, 1]nums2 = [2, 2], return [2, 2].
Note:
  • Each element in the result should appear as many times as it shows in both arrays.
  • The result can be in any order.
Follow up:
  • What if the given array is already sorted? How would you optimize your algorithm?
  • What if nums1's size is small compared to num2's size? Which algorithm is better?
  • What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?
http://bookshadow.com/weblog/2016/05/21/leetcode-intersection-of-two-arrays-ii/

  • 如果数组已经排好序,怎样优化你的算法?
  • 如果nums1的长度<nums2的长度?哪一种算法更好?
  • 如果nums2的元素存储在磁盘上,并且内存大小有限,不足以将其一次性的加载到内存中。此时应当怎样做?
  • X. 解法I 排序(Sort)+双指针(Two Pointers)
    http://vegetablefinn.github.io/2016/05/21/350-Intersection-of-Two-Arrays-II/
    If the arrays are sorted, then we can use two points.
        public int[] intersect(int[] nums1, int[] nums2) {
            //先排序
            Arrays.sort(nums1);
            Arrays.sort(nums2);
            //确定head
            int nums1Head = 0;
            int nums2Head = 0;
            //确定输出数组
            List<Integer> resultList = new ArrayList<Integer>();
            while(nums1Head < nums1.length && nums2Head < nums2.length ){
              if(nums1[nums1Head] == nums2[nums2Head]){
                resultList.add(nums1[nums1Head]);
                nums1Head++;
                nums2Head++;
                continue;
              }else if(nums1[nums1Head] < nums2[nums2Head]){
                nums1Head++;
                continue;
              }else if(nums1[nums1Head] > nums2[nums2Head]){
                nums2Head++;
                continue;
              }
            }
            int result[] = new int[resultList.size()];
            for (int i=0; i < result.length; i++)
            {
                result[i] = resultList.get(i).intValue();
            }
            return result;
        }

    解法II Counter计数
    该解法不需要将nums2一次性加载到内存中
    def intersect(self, nums1, nums2): """ :type nums1: List[int] :type nums2: List[int] :rtype: List[int] """ if len(nums1) > len(nums2): nums1, nums2 = nums2, nums1 c = collections.Counter(nums1) ans = [] for x in nums2: if c[x] > 0: ans += x, c[x] -= 1 return ans

    Map<Integer, Long> map = Arrays.stream(nums2).boxed().collect
    (Collectors.groupingBy(e->e, Collectors.counting())); return Arrays.stream(nums1).filter(e ->{ if(!map.containsKey(e)) return false; map.put(e, map.get(e) - 1); if(map.get(e) == 0) map.remove(e); return true; }).toArray();

    [JAVA] What's the best way you think to convert List<Integer> to a normal int array?
    int[] array = list.stream().mapToInt(i->i).toArray();
    
    list.stream().mapToInt(Integer :: intValue).toArray()
    
    X. Follow up
    https://leetcode.com/discuss/103969/solution-to-3rd-follow-up-question
    What if elements of nums2 are stored on disk, and the memory is
    limited such that you cannot load all elements into the memory at
    once?
    • If only nums2 cannot fit in memory, put all elements of nums1 into a HashMap, read chunks of array that fit into the memory, and record the intersections.
    • If both nums1 and nums2 are so huge that neither fit into the memory, sort them individually (external sort), then read 2 elements from each array at a time in memory, record intersections.
    I think the second part of the solution is impractical, if you read 2 elements at a time, this procedure will take forever. In principle, we want minimize the number of disk access during the run-time.
    An improvement can be sort them using external sort, read (let's say) 2G of each into memory and then using the 2 pointer technique, then read 2G more from the array that has been exhausted. Repeat this until no more data to read from disk.
    But I am not sure this solution is good enough for an interview setting. Maybe the interviewer is expecting some solution using Map-Reduce paradigm.
    http://faculty.simpson.edu/lydia.sinapova/www/cmsc250/LN250_Weiss/L17-ExternalSortEX2.htm
    http://massivealgorithms.blogspot.com/2014/07/exceptional-code-external-sorting-for.html
    http://buttercola.blogspot.com/2016/06/leetcode-intersection-of-two-arrays-ii.html

    • What if the given array is already sorted? How would you optimize your algorithm?
      • Solution 1, i.e., sorting,  would be better since it does not need extra memory. 
    • What if nums1's size is small compared to num2's size? Which algorithm is better?
      • If two arrays are sorted, we could use binary search, i.e., for each element in the shorter array, search in the longer one. So the overall time complexity is O(nlogm), where n is the length of the shorter array, and m is the length of the longer array. Note that this is better than Solution 1 since the time complexity is O(n + m) in the worst case. 
    • What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?
      • If the two arrays have relatively the same length, we can use external sort to sort out the two arrays in the disk. Then load chunks of each array into the memory and compare, by using the method 1. 
      • If one array is too short while the other is long, in this case, if memory is limited and nums2 is stored in disk, partition it and send portions of nums2 piece by piece. keep a pointer for nums1 indicating the current position, and it should be working fine~
      • Another method is, store the larger array into disk, and for each element in the shorter array, use "Exponential Search" and search in the longer array. 
    https://helloyuantechblog.wordpress.com/2016/06/20/leetcode-350-intersection-of-two-arrays-ii-easy/
        vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
            //sort(nums1.begin(), nums1.end()); not needed
            sort(nums2.begin(), nums2.end());
            vector<int> result;
            int len1 = nums1.size(), len2 = nums2.size();
            for(int i = 0; i < nums1.size(); i++) { if (i > 0 && nums1[i] == nums1[i-1]) {
                    continue;
                }
                int j = binarySearch(nums2, nums1[i]);
                if (j != -1) {
                    result.push_back(nums1[i]);
                    while(i + 1 < len1 && nums1[i+1] == nums1[i] && j + 1 < len2 && nums2[j+1] == nums2[j]) {
                        result.push_back(nums1[i]);
                        i++; j++;
                    }
                }
            }
            return result;
        }
        int binarySearch(vector<int>& nums, int target) {
            if (nums.empty()) {
                return -1;
            }
            int start = 0, end = nums.size()-1, mid;
            while(start + 1 < end) {
                mid = start + (end - start) / 2;
                if (nums[mid] < target) {
                    start = mid + 1;
                } else {
                    end = mid;
                }
            }
            if (nums[start] == target) {
                return start;
            }
            if (nums[end] == target) {
                return end;
            }
            return -1;
        }
    https://discuss.leetcode.com/topic/49329/all-the-follow-up-solution
    1.What if the given array is already sorted? How would you optimize your algorithm?
    class Solution {
    public:
        vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
            vector<int> res;
            if(nums1.empty() || nums2.empty())
                return res;
            for(int i=0,j=0; i<nums1.size() && j<nums2.size(); ) {
                if(nums1[i] == nums2[j]) {
                    res.push_back(nums2[j]);
                    ++j;
                    ++i;
                }else if(nums1[i] < nums2[j]) {
                    ++i;
                }else if(nums1[i] > nums2[j]) {
                    ++j;
                }
            }
            return res;
        }
    };
    
    2.What if nums1's size is small compared to nums2's size? Which algorithm is better?
    class Solution {
    public:
        vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
            unordered_map<int, int> mymap;
            vector<int> res;
            for(auto it : nums1) {
                mymap[it]++;
            }
            for(auto it : nums2) {
                if(--mymap[it] >= 0) {
                    res.push_back(it);
                }
            }
            return res;
    
        }
    };



    No comments:

    Post a Comment

    Labels

    GeeksforGeeks (959) Algorithm (811) LeetCode (637) to-do (597) Review (339) Classic Algorithm (334) Classic Interview (299) Dynamic Programming (263) Google Interview (233) LeetCode - Review (228) 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) Trie (33) prismoskills (33) Segment Tree (32) Space Optimization (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) 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) 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) 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) 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) 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