Showing posts with label Company-Airbnb. Show all posts
Showing posts with label Company-Airbnb. Show all posts

Implement Queue with Limited Size of Array - Airbnb


https://leetcode.com/discuss/interview-question/156969/Implement-Queue-using-fixed-size-array/


Assume in your programming language you only have a fixed size array of length 5. Implement a queue datastructure that can get unlimitted number of elements.

Assumption: The arrays hold 5 elements, each of which is the size of a pointer.
Our first level of abstraction is to use our stupid arrays to make a less stupid array, which has access in log(N) time.
Once we have an Array, with all of the capabilities of your typical array, implementing a queue, stack, deque, whatever, will be the normal way of doing so.
class Array {
    Array(int size) {
        this->size = size;
        
        depth = max(0, log5(size)-1); // Depth 0 is for size <= 5
        build_array(depth, head);
    }

    // Will hold 5^depth elements, which is likely greater than size, but you know
    // what they say about premature optimization
    void build_array(int depth, stupid_array arr) {
        if (depth == 0) return;
        for i = 0 to 4 {
            arr[i] = new stupid_array;
            build_array(depth-1, arr[i]);
        }
    }

    (stupid_array, int) find_index(int idx)
    {
        return find_index(0, idx, head);
    }

    (stupid_array, int) find_index(int d, int idx, stupid_array arr)
    {
        int net_depth = depth - d;
        int subarray_size = pow(5, net_depth-1);
        if (subarray_size == 1) {
            return (head, idx);
        }

        return find_index(d+1, idx % subarray_size, arr[idx / subarray_size]);
    }

    insert(void * value, int idx) {
        stupid_array subarr;
        int subidx;
        (subarr, subidx) =  find_index(idx);
        subarr[subidx] = value;
    } 

    void * get(int idx) {
        stupid_array subarr;
        int subidx;
        (subarr, subidx) =  find_index(idx);
        return subarr[subidx];
    }

    stupid_array head;
    int size, depth;
}
Now that we have an array that has a normal interface, and which has reasonably fast access of O(log(N)), we can easily create a queue
https://github.com/allaboutjst/airbnb
Implement a queue with a number of arrays, in which each array has fixed size.
2.Implement Queue with limited size of array: 使用double linkedlist
2. Implement Queue with limited size of array:方法二 ListNode with fixed size of array
    public class QueueWithFixedArray {
        private int fixedSize;

        private int count;
        private int head;
        private int tail;
        private List<Object> headList;
        private List<Object> tailList;

        public QueueWithFixedArray(int fixedSize) {
            this.fixedSize = fixedSize;
            this.count = 0;
            this.head = 0;
            this.tail = 0;
            this.headList = new ArrayList<>();
            this.tailList = this.headList;
        }

        public void offer(int num) {
            if (tail == fixedSize - 1) {
                List<Object> newList = new ArrayList<>();
                newList.add(num);
                tailList.add(newList);
                tailList = (List<Object>) tailList.get(tail);
                tail = 0;
            } else {
                tailList.add(num);
            }
            count++;
            tail++;
        }

        public Integer poll() {
            if (count == 0) {
                return null;
            }

            int num = (int) headList.get(head);
            head++;
            count--;

            if (head == fixedSize - 1) {
                List<Object> newList = (List<Object>) headList.get(head);
                headList.clear();
                headList = newList;
                head = 0;
            }

            return num;
        }

        public int size() {
            return count;
        }
    }




TODO:
http://fabian-kostadinov.github.io/2014/11/25/implementing-a-fixed-length-fifo-queue-in-java/

Minimum Vertices to Traverse Directed Graph - Airbnb


https://leetcode.com/discuss/interview-question/124861/digraph-cover-all-vertices-with-the-least-number-of-vertices
Given a directed graph G (can contain sub graphs and cycles), find the minimum number of vertices from which all nodes are reachable.
For example:
Nodes:
    0, 1, 2, 3, 4, 5

Edges:
    1 <- 0
    0 <- 1 <- 2
    3 <- 1 <- 2
    2 <- 5
    4 <- 5

Representation:
    --> 4
  /
 5 --> 2 --> 1 <--> 0
              \
                --> 3
Matrix:
g = [[1, 1, 0, 0, 0, 0],
     [1, 1, 1, 0, 0, 0],
     [0, 0, 1, 0, 0, 1],
     [0, 1, 0, 1, 0, 0],
     [0, 0, 0, 0, 1, 1],
     [0, 0, 0, 0, 0, 1]]
Return 1 (node number 5). From node number 5 all other nodes are reachable.
If we remove edge 2 <- 5, the result is 2, because we need at least nodes number 5 and 2 to visit all nodes.
Representation of the graph if we remove the edge between nodes 2 and 5.
 5 --> 4

 2 --> 1 <--> 0
        \
          --> 3
I attempted to solve this problem by finding for every node j, and array of all nodes reachable from j. Basically, I did a DFS on every node, and the result looks like this:
{
    0: [True, True, False, True, False, False],
    1: [True, True, False, True, False, False],
    2: [True, True, True, True, False, False],
    3: [False, False, False, True, False, False],
    4: [False, False, False, False, True, False],
    5: [True, True, True, True, True, True]
}
This means that from node 0, I can reach nodes number 0, 1, 3. From node 1 I can reach nodes 0, 1 and 3. From node 5, I can reach all nodes.
Is this a good approach to solve the problem? Having this dictionary, how can I find the smallest group of nodes from which I can reach all nodes?

It's the number of strongly-connected components that aren't reachable from any outside vertex.

What if you found all vertices with no incoming edges and started BFS from those vertices? If you run out of vertices with no incoming edges, then that means the rest of the vertices are either in a cycle or completely separated.
Each time you have to search from a new vertex, you would add 1 to your overall total vertices required.
Effectively you would run DFS on each "connected component", making the overall runtime O(E + V) to find the number of necessary vertices to explore all other vertices.
Is this a viable solution or am I mistaken?
https://github.com/allaboutjst/airbnb/blob/master/README.md
Given a directed grapjh, represented in a two dimension array, output a list of points that can be used to travese every points with the least number of visited vertices.
        private void search(Set<Integer> res, Map<Integer, Set<Integer>> nodes, int cur, int start,
                            Set<Integer> visited, Set<Integer> currVisited) {
            currVisited.add(cur);
            visited.add(cur);
            for (int next : nodes.get(cur)) {
                if (res.contains(next) && next != start) {
                    res.remove(next);
                }
                if (!currVisited.contains(next)) {
                    search(res, nodes, next, start, visited, currVisited);
                }
            }
        }

        public List<Integer> getMin(int[][] edges, int n) {
            Map<Integer, Set<Integer>> nodes = new HashMap<>();
            for (int i = 0; i < n; i++) {
                nodes.put(i, new HashSet<>());
            }
            for (int[] edge : edges) {
                nodes.get(edge[0]).add(edge[1]);
            }

            Set<Integer> visited = new HashSet<>();
            Set<Integer> res = new HashSet<>();
            for (int i = 0; i < n; i++) {
                if (!visited.contains(i)) {
                    res.add(i);
                    visited.add(i);
                    search(res, nodes, i, i, visited, new HashSet<>());
                }
            }

            return new ArrayList<>(res);
        }

https://rextester.com/discussion/NJE76317/Minimum-Vertices-to-Traverse-Directed-Graph


Finding Ocean - Airbnb


https://github.com/allaboutjst/airbnb/blob/master/README.md
Given: An array of strings where L indicates land and W indicates water, and a coordinate marking a starting point in the middle of the ocean.
Challenge: Find and mark the ocean in the map by changing appropriate Ws to Os.
  • An ocean coordinate is defined to be the initial coordinate if a W, and
  • any coordinate directly adjacent to any other ocean coordinate.
void findOcean(String[] map, int row, int column);
String[] map = new String[]{
 "WWWLLLW",
 "WWLLLWW",
 "WLLLLWW"
};
printMap(map);
STDOUT:
WWWLLLW
WWLLLWW
WLLLLWW
findOcean(map, 0, 1);
printMap(map);
STDOUT:
OOOLLLW
OOLLLWW
OLLLLWW

https://github.com/allaboutjst/airbnb/blob/master/src/main/java/finding_ocean/FindingOcean.java
        public void floodFill(char[][] board, int i, int j, char oldColor, char newColor) {
            if (board[i][j] != oldColor || board[i][j] == newColor) {
                return;
            }

            Queue<Integer> queue = new LinkedList<>();
            queue.add(i * board[0].length + j);
            board[i][j] = newColor;

            while (!queue.isEmpty()) {
                int pos = queue.poll();
                int m = pos / board[0].length;
                int n = pos % board[0].length;

                if (m + 1 < board.length && board[m + 1][n] == oldColor) {
                    queue.add((m + 1) * board[0].length + n);
                    board[m + 1][n] = newColor;
                }
                if (m - 1 >= 0 && board[m - 1][n] == oldColor) {
                    queue.add((m - 1) * board[0].length + n);
                    board[m - 1][n] = newColor;
                }
                if (n + 1 < board[0].length && board[m][n + 1] == oldColor) {
                    queue.add(m * board[0].length + n + 1);
                    board[m][n + 1] = newColor;
                }
                if (n - 1 >= 0 && board[m][n - 1] == oldColor) {
                    queue.add(m * board[0].length + n - 1);
                    board[m][n - 1] = newColor;
                }
            }
        }


Menu Combination Sum - Airbnb


https://github.com/allaboutjst/airbnb/blob/master/README.md
Given a menu (list of items prices), find all possible combinations of items that sum a particular value K. (A variation of the typical 2sum/Nsum questions).

        private void search(List<List<Double>> res, int[] centsPrices, int start, int centsTarget,
                            List<Double> curCombo, double[] prices) {
            if (centsTarget == 0) {
                res.add(new ArrayList<>(curCombo));
                return;
            }

            for (int i = start; i < centsPrices.length; i++) {
                if (i > start && centsPrices[i] == centsPrices[i - 1]) {
                    continue;
                }
                if (centsPrices[i] > centsTarget) {
                    break;
                }
                curCombo.add(prices[i]);
                search(res, centsPrices, i + 1, centsTarget - centsPrices[i], curCombo, prices);
                curCombo.remove(curCombo.size() - 1);
            }
        }

        public List<List<Double>> getCombos(double[] prices, double target) {
            List<List<Double>> res = new ArrayList<>();
            if (prices == null || prices.length == 0 || target <= 0) {
                return res;
            }

            int centsTarget = (int) Math.round(target * 100);
            Arrays.sort(prices);
            int[] centsPrices = new int[prices.length];
            for (int i = 0; i < prices.length; i++) {
                centsPrices[i] = (int) Math.round(prices[i] * 100);
            }

            search(res, centsPrices, 0, centsTarget, new ArrayList<>(), prices);
            return res;
        }


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