Sunday, November 29, 2015

Meeting Room Misc


http://yuanhsh.iteye.com/blog/2196848
At a bus-station, you have time-table for buses arrival and departure. You need to find the minimum number of platforms so that all the buses can be accommodated as per their schedule.
Example: Time table is like below:
Bus         Arrival         Departure 
BusA        0900 hrs        0930 hrs
BusB        0915 hrs        1300 hrs
BusC        1030 hrs        1100 hrs
BusD        1045 hrs        1145 hrs

Then the answer must be 3. Otherwise the bus-station will not be able to accommodate all the buses.
  1. int findPlatform(int arr[], int dep[], int n)  
  2. {  
  3.    // Sort arrival and departure arrays  
  4.    sort(arr, arr+n);  
  5.    sort(dep, dep+n);  
  6.    
  7.    // plat_needed indicates number of platforms needed at a time  
  8.    int plat_needed = 1, result = 1;  
  9.    int i = 1, j = 0;  
  10.    
  11.    // Similar to merge in merge sort to process all events in sorted order  
  12.    while (i < n && j < n)  
  13.    {  
  14.       // If next event in sorted order is arrival, increment count of  
  15.       // platforms needed  
  16.       if (arr[i] < dep[j])  
  17.       {  
  18.           plat_needed++;  
  19.           i++;  
  20.           if (plat_needed > result)  // Update result if needed  
  21.               result = plat_needed;  
  22.       }  
  23.       else // Else decrement count of platforms needed  
  24.       {  
  25.           plat_needed--;  
  26.           j++;  
  27.       }  
  28.    }  
  29.    
  30.    return result;  
  31. }

  1. int min_rooms(vector<Interval>& meetings) {  
  2.     vector<int> times;  
  3.     for(auto m : meetings) {  
  4.         times.push_back(m.begin);  
  5.         times.push_back(-m.end);  
  6.     }     
  7.     sort(times.begin(), times.end(), [](int a, int b){  
  8.         return abs(a) == abs(b) ? a < b : abs(a) < abs(b);  
  9.     });     
  10.     int ret = 0, cur = 0;  
  11.     for(auto t : times) {  
  12.         if(t >= 0) ret = max(ret, ++cur);  
  13.         else --cur;  
  14.     }  
  15.     return ret;  
  16. }  

The follow up question:
You have a number of meetings (with their start and end times). You need to schedule them using the minimum number of rooms. Return the list of meetings in every room.

First we need to realize that randomly schedule meetings to available rooms won’t give you an optimal solution. For example, suppose there are 4 meetings, the (start, end) time are (1, 3), (1, 5), (6, 7), (4, 7). When (1, 3) comes, assign to room A, then (1, 5) comes, assign to room B, then (6, 7) comes, assign to room A as it is available, then (4, 7) comes, both room A and B are not available so we have to assign a new room C for it. However, a better solution is two rooms, room A for meeting (1, 3) (4, 7) and room B for meeting (1, 5) (6, 7).
However, the optimal solution solution is not far from it. If we first sort all the meeting by the start time, then we could just assign them one by one and to the first available room.
The reason is simple, when considering a meeting from s[i] to e[i], as there is no unschedule meeting before s[i], by putting the (s[i], e[i]) meeting in any available room (if there is one) would leads to the same results.
So the code looks like this.
  1. void arrange(vector<pair> meeting) { // the pair contains the start and end time.  
  2.   sort(meeting.begin(), meeting.end());  
  3.   vector<vector<pair>> room; // for each room, there is a list of meetings.  
  4.   for (auto m : meeting) {  
  5.     bool found = False;  
  6.     for (auto& r : room) {  
  7.       if (m.first >= r.back().second) // we can arrange the meeting in the room  
  8.         r.push_back(m);  
  9.         found = True;  
  10.         break;  
  11.       }  
  12.     }  
  13.     if (!found) {  
  14.       room.push_back({m});  
  15.     }  
  16.   }  
  17.   cout << "Requires " << room.size() << " rooms" << endl;  
  18.   for (int i = 0; i < room.size(); ++i) {  
  19.     cout << "Room " << i << endl;  
  20.     for (auto m : room[i]) {  
  21.       cout << "Meeting: " << m.first << " " << m.second << endl;  
  22.     }  
  23.   }  

https://hellosmallworld123.wordpress.com/2014/05/30/arranging-the-meeting-room/
If you have only one room, what is the maximum number of meetings you can scheduled into that room.
Solution:
1. sort the meetings by finishing time, this is because we greedily choose the meeting that finishes first.
2. go through all the meetings in order of finishing time, schedule the meeting into the room if the room is not occupied at its start time, and increase the count by one.
3. no of count will be the max number of meetings you can schedule into the room.
We need to prove that the greedy algorithm is correct (choosing the meeting that finishes first can result in a optimal solution) assume there is another schedule S’ that schedules more meetings (k + 1) then the solution S (k solutions). Then at some point the S’ must scheduled some meeting that tm’ ends before the tm scheduled by S. But as we know that since S scheduled meeting that finishes first so the mth meeting must finishes no later than mth scheduled by S’. which is a contradiction.
Feasible problem:
Given a serial of jobs that contains a processTime and a deadLine, check if you can schedule them so that all the Jobs are finished before deadline.
1.Sort the jobs by deadline.
2. Schedule the job by the order of deadline, see if any job can not finish before deadline, if so, it is not possible, otherwise, the problem is feasible.
Minimum Lateness problem:
Assume we have a serial of jobs that contains a processTime and a deadLine, the lateness is the time between the finish time and the deadline of the job, if the job is finished before the deadline, then the lateness is 0. How do you schedule the job to minimize the total lateness?
1. Sort the jobs in deadline
2. step through all the elements in the order of the deadline, pick the earliest deadline and sums up the lateness if exist.
3. return the lateness
Third follow up question, what if each meeting has a priority, how can we schedule a set of meeting that has the largest priority sum given one meeting room?
1. Sort the meetings according to the finish time (like the first question)
2. Try schedule meeting starting from the first one greedily and record the total priority, mark those meetings that has been scheduled
3. Pass through all the meetings and schedule any meeting that has not been marked in previous schedules. record the max total priority



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