Tuesday, December 15, 2015

LeetCode [317] Shortest Distance from All Buildings


http://algobox.org/shortest-distance-from-all-buildings/
You want to build a house on an empty land which reaches all buildings in the shortest amount of distance. You can only move up, down, left and right. You are given a 2D grid of values 01 or 2, where:
  • Each 0 marks an empty land which you can pass by freely.
  • Each 1 marks a building which you cannot pass through.
  • Each 2 marks an obstacle which you cannot pass through.
Example:
Given three buildings at (0,0)(0,4)(2,2), and an obstacle at (0,2):
The point (1,2) is an ideal empty land to build a house, as the total travel distance of 3+3+1=7 is minimal.
Return 7.
Note:
There will be at least one building. If it is not possible to build such house according to the above rules, return -1.
https://leetcode.com/discuss/76018/share-a-java-implement
https://discuss.leetcode.com/topic/31668/my-bfs-java-solution
Short version: BFS from every building, calculate the distances and find the minimum distance in the end.
Key optimization : we do not go into a land, if it is not accessible by at least one of previous buildings.

The shortest distance from any empty land to a building can be calculated by BFS starting from the building in O(mn)time. Therefore the we can calculate distance from all buildings to empty lands by t rounds of BFS starting from each building. t is the total number of buildings.
In each round, we need maintain two values for every empty land: the distance and the accessibility.
  • dist[i][j] is the empty land (i, j) to all the buildings.
  • grid[i][j] is reused as the accessibility.
What is accessibility? It is the number of buildings that are accessible from the empty land.
In each round of BFS we can maintain these two values. In the end we just need to find the minimum value of dist[i][j] where the accessibility equals t.
One interesting point is that the grid[i][j] can also be used to check if (i, j) is visited in this round. At round k(zero based), those has grid[i][j] == k is the empty land unvisited, visited land will have grid[i][j] == k + 1. Here comes the interesting part. One may ask what if grid[i][j] < k? Answer is we do not go into the lands withgrid[i][j] < k as if it is an obstacle. Why can we do that? Because grid[i][j] < k means it is not accessible by at least one of the buildings in previous rounds. Which means not only this land is not our answer, all the lands accessible from (i, j) is also not our answer.
This might be why it runs faster than many implements. The Java version runs in 13 ms.
int[] dx = {1, 0, -1, 0}, dy = {0, 1, 0, -1}; public int shortestDistance(int[][] grid) { int m = grid.length, n = grid[0].length; int[][] dist = new int[m][n]; // Initialize building list and accessibility matrix `grid` List<Tuple> buildings = new ArrayList<>(); for (int i = 0; i < m; ++i) for (int j = 0; j < n; ++j) { if (grid[i][j] == 1) buildings.add(new Tuple(i, j, 0)); grid[i][j] = -grid[i][j]; } // BFS from every building for (int k = 0; k < buildings.size(); ++k) bfs(buildings.get(k), k, dist, grid, m, n); // Find the minimum distance int ans = -1; for (int i = 0; i < m; ++i) for (int j = 0; j < n; ++j) if (grid[i][j] == buildings.size() && (ans < 0 || dist[i][j] < ans)) ans = dist[i][j]; return ans; } public void bfs(Tuple root, int k, int[][] dist, int[][] grid, int m, int n) { Queue<Tuple> q = new ArrayDeque<>(); q.add(root); while (!q.isEmpty()) { Tuple b = q.poll(); dist[b.y][b.x] += b.dist; for (int i = 0; i < 4; ++i) { int x = b.x + dx[i], y = b.y + dy[i]; if (y >= 0 && x >= 0 && y < m && x < n && grid[y][x] == k) { grid[y][x] = k + 1; q.add(new Tuple(y, x, b.dist + 1)); } } } } class Tuple { public int y; public int x; public int dist; public Tuple(int y, int x, int dist) { this.y = y; this.x = x; this.dist = dist; } }
https://leetcode.com/discuss/74999/java-solution-with-explanation-and-time-complexity-analysis
Traverse the matrix. For each building, use BFS to compute the shortest distance from each '0' to this building. After we do this for all the buildings, we can get the sum of shortest distance from every '0' to all reachable buildings. This value is stored in 'distance[][]'. For example, if grid[2][2] == 0, distance[2][2] is the sum of shortest distance from this block to all reachable buildings. Time complexity: O(number of 1)O(number of 0) ~ O(m^2n^2)
We also count how many building each '0' can be reached. It is stored in reach[][]. This can be done during the BFS. We also need to count how many total buildings are there in the matrix, which is stored in 'buildingNum'.
Finally, we can traverse the distance[][] matrix to get the point having shortest distance to all buildings. O(m*n)
The total time complexity will be O(m^2*n^2), which is quite high!.

Isn't the time complexity O(#buildings * m^2 * n^2)? BFS's time complexity is O(|V||E|). Here we have mn vertices and the edges are proportional to m*n, so every BFS is O(m^2 * n^2). We do a BFS for every building so the total time complexity is O(#buildings * m^2 * n^2)?

The time complexity for BFS/DFS is O(|V|+|E|), not O(|V||E|). In this problem, every vertex has up to 4 edges (left, right, up, down), so |E| ~ 4|V|. Thus, you have overall O(|V|) = O(mn) for a BFS. This has been proven for all sparse graphs like this problem. Now, we do a BFS for each building, so the overall complexity is O(#buildings*(mn)). In worst case, every vertex is a building. So the number of buildings is also upper bounded by O(mn), and thus you have O((mn)(mn)) = O(m^2n^2). This is a very loose bound since when every vertex is a building, we don't even need to do a BFS (nowhere to go).

public int shortestDistance(int[][] grid) { if (grid == null || grid[0].length == 0) return 0; final int[] shift = new int[] {0, 1, 0, -1, 0}; int row = grid.length, col = grid[0].length; int[][] distance = new int[row][col]; int[][] reach = new int[row][col]; int buildingNum = 0; for (int i = 0; i < row; i++) { for (int j =0; j < col; j++) { if (grid[i][j] == 1) { buildingNum++; Queue<int[]> myQueue = new LinkedList<int[]>(); myQueue.offer(new int[] {i,j}); boolean[][] isVisited = new boolean[row][col]; int level = 1; while (!myQueue.isEmpty()) { int qSize = myQueue.size(); for (int q = 0; q < qSize; q++) { int[] curr = myQueue.poll(); for (int k = 0; k < 4; k++) { int nextRow = curr[0] + shift[k]; int nextCol = curr[1] + shift[k + 1];
// if reach[nextRow][nextCol] < buildingNum, ignore it.
if (nextRow >= 0 && nextRow < row && nextCol >= 0 && nextCol < col && grid[nextRow][nextCol] == 0 && !isVisited[nextRow][nextCol]) { //The shortest distance from [nextRow][nextCol] to thic building // is 'level'. distance[nextRow][nextCol] += level; reach[nextRow][nextCol]++; isVisited[nextRow][nextCol] = true; myQueue.offer(new int[] {nextRow, nextCol}); } } } level++; } } } } int shortest = Integer.MAX_VALUE; for (int i = 0; i < row; i++) { for (int j = 0; j < col; j++) { if (grid[i][j] == 0 && reach[i][j] == buildingNum) { shortest = Math.min(shortest, distance[i][j]); } } } return shortest == Integer.MAX_VALUE ? -1 : shortest; }

http://www.jianshu.com/p/c38f93f6431f
What's the benefit?
这道题目让我想起了 multi-end BFS
就是从多个building同时出发,一起遍历。
但是问题在于,如何标志,这个 empty area 被多个Building访问过后的状态?

I think a better solution is using BFS to walk from all buildings together and stop in the middle when all of them meet, instead of completing all spots. However, my run time is just 58ms. I think it is because I used custom class.
your solution seems to be fine at the first glance, but just want to be clear, you did not stop in the middle when all of them meet at the first time right? You stop when the accumulated reachable distance is no more less than the minimum distance. Otherwise, test case like [[1,0,1],[2,0,2],[2,1,2]] would be wrong. Also, you might push land at the same locations multiple times to the queue.
static final int[] s={-1,0,1,0,-1};

class Move{
    int x;
    int steps;
    boolean[][] visited;
    Move(int p, int s){
        x=p;
        steps=s;
    }
}

public int shortestDistance(int[][] grid) {
    if(grid==null || grid.length==0 || grid[0].length==0) return 0;
    int m = grid.length, n=grid[0].length, total=0, min=Integer.MAX_VALUE;
    int[][] distance= new int[m][n];
    int[][] reach= new int[m][n];
    Deque<Move> queue = new ArrayDeque<>();
    for(int i=0; i<m; i++){
        for(int j=0; j<n; j++){
            if(grid[i][j]==1){
                Move mv = new Move(i*n+j, 0);
                mv.visited = new boolean[m][n];
                mv.visited[i][j]=true;
                queue.offer(mv);
                total++;
            }
        }
    }
    while(!queue.isEmpty()){
        Move mv = queue.poll();
        int x = mv.x/n, y=mv.x%n;
        for(int i=0; i<4; i++){
            int p = x+s[i], q=y+s[i+1];
            if(p>=0 && p<m && q>=0 && q<n && !mv.visited[p][q] && grid[p][q]==0){
                Move newMv = new Move(p*n+q, mv.steps+1);
                newMv.visited=mv.visited;
                newMv.visited[p][q]=true;
                distance[p][q]+=mv.steps+1;
                if(distance[p][q]<min) {
                    if(++reach[p][q]==total) min = distance[p][q];
                    queue.offer(newMv);
                }
            }
        }
    }
    return (min==Integer.MAX_VALUE)?-1: min;
}
http://www.cnblogs.com/yrbbest/p/5068730.html
https://leetcode.com/discuss/74380/my-bfs-java-solution
    private final int[][] directions = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
    
    public int shortestDistance(int[][] grid) {
        if(grid == null || grid.length == 0) {
            return Integer.MAX_VALUE;
        }
        int rowNum = grid.length;
        int colNum = grid[0].length;
        int[][] distance = new int[rowNum][colNum];
        int[][] canReachBuildings = new int[rowNum][colNum];
        int buildingNum = 0; 
        
        for(int i = 0; i < rowNum; i++) {
            for(int j = 0; j < colNum; j++) {
                if(grid[i][j] != 0) {
                    distance[i][j] = Integer.MAX_VALUE;
                }
                if(grid[i][j] == 1) {     // find out all buildings
                    buildingNum++;
                    updateDistance(grid, distance, canReachBuildings, i, j);
                }
            }
        }
        
        int min = Integer.MAX_VALUE;
        for(int i = 0; i < rowNum; i++) {
            for(int j = 0; j < colNum; j++) {
                if(canReachBuildings[i][j] == buildingNum) {
                    min = Math.min(distance[i][j], min);
                }
            }
        }
        
        return min == Integer.MAX_VALUE ? -1 : min;
    }
    
    private void updateDistance(int[][] grid, int[][] distance, int[][] canReachBuildings, int row, int col) {
        Queue<int[]> queue = new LinkedList<>();
        queue.offer(new int[]{row, col});
        boolean[][] visited = new boolean[grid.length][grid[0].length];
        visited[row][col] = true;
        int dist = 0;
        int curLevel = 1;
        int nextLevel = 0;
        
        while(!queue.isEmpty()) {
            int[] position = queue.poll();
            distance[position[0]][position[1]] += dist;
            curLevel--;
            for(int[] direction : directions) {
                int x = position[0] + direction[0];
                int y = position[1] + direction[1];
                if(x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] != 0) {
                    continue;
                }
                if(!visited[x][y]) {
                    queue.offer(new int[]{x, y}); 
                    nextLevel++;
                    visited[x][y] = true;
                    canReachBuildings[x][y]++;
                }
            }
            if(curLevel == 0) {
                curLevel = nextLevel;
                nextLevel = 0;
                dist++;
            }
        }
    }
http://buttercola.blogspot.com/2016/01/leetcode-shortest-distance-from-all.html
A BFS problem. Search from each building and calculate the distance to the building. One thing to note is an empty land must be reachable by all buildings. To achieve this, maintain an array of counters. Each time we reach an empty land from a building, increase the counter. Finally, a reachable point must have the counter equaling to the number of buildings. 
    public int shortestDistance(int[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }
         
        int m = grid.length;
        int n = grid[0].length;
         
        int[][] dist = new int[m][n];
        int[][] reach = new int[m][n];
        // step 1: BFS and calcualte the min dist from each building
        int numBuildings = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    boolean[][] visited = new boolean[m][n]; // no need declare here
                    Queue<Integer> queue = new LinkedList<>();
                    shortestDistanceHelper(i, j, 0, dist, reach, grid, visited, queue);
                    numBuildings++;
                }
            }
        }
         
        // step 2: caluclate the minimum distance
        int minDist = Integer.MAX_VALUE;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 0 && reach[i][j] == numBuildings) {
                    minDist = Math.min(minDist, dist[i][j]);
                }
            }
        }
         
        return minDist == Integer.MAX_VALUE ? -1 : minDist;
    }
     
    private void shortestDistanceHelper(int x, int y, int currDist,
                                        int[][] dist, int[][] reach, int[][] grid,
                                        boolean[][] visited, Queue<Integer> queue) {
        fill(x, y, x, y, currDist, dist, reach, grid, visited, queue);
         
        int m = grid.length;
        int n = grid[0].length;
         
        while (!queue.isEmpty()) {
            int size = queue.size();
            currDist++;
            for (int sz = 0; sz < size; sz++) {
                int cord = queue.poll();
                int i = cord / n;
                int j = cord % n;
                 
                fill(x, y, i - 1, j, currDist, dist, reach, grid, visited, queue);
                fill(x, y, i + 1, j, currDist, dist, reach, grid, visited, queue);
                fill(x, y, i, j - 1, currDist, dist, reach, grid, visited, queue);
                fill(x, y, i, j + 1, currDist, dist, reach, grid, visited, queue);
            }
        }
    }
     
    private void fill(int origX, int origY, int x, int y, int currDist,
                      int[][] dist, int[][] reach, 
                      int[][] grid, boolean[][] visited, Queue<Integer> queue) {
         
        int m = dist.length;
        int n = dist[0].length;
        if (x < 0 || x >= m || y < 0 || y >= n || visited[x][y]) {
            return;
        }
         
        if ((x != origX || y != origY) && grid[x][y] != 0) {
            return;
        }
         
        visited[x][y] = true;
         
        dist[x][y] += currDist;
        reach[x][y]++;
         
        queue.offer(x * n + y);
    }
https://leetcode.com/discuss/92508/11ms-simple-java-soluton-that-beats-100%25-with-explanation
public int shortestDistance(int[][] grid) { int[][] d = new int[grid.length][grid[0].length];//sum of distance from all buildings int[][] cc = new int[grid.length][grid[0].length];//reachable by how many buildings int count = 0; //find how many buildings for(int i = 0;i<grid.length;i++){ for(int j = 0;j<grid[0].length;j++){ if(grid[i][j] == 1) { count++; } } } for(int i = 0;i<grid.length;i++){ for(int j = 0;j<grid[0].length;j++){ if(grid[i][j] == 1) { //bfs each building, keep on updating empty spot's distance value boolean x = bfs(grid, i, j, d, new boolean[grid.length][grid[0].length], count, cc); if(!x) return -1;//if this building can not reach all other buildings return -1 } } } //find the smallest distance, need be an empty spot, and a spot can be reached by all buildings int res = Integer.MAX_VALUE; for(int i = 0;i<grid.length;i++){ for(int j = 0;j<grid[0].length;j++){ if(d[i][j] != 0 && cc[i][j] == count) res = Math.min(res, d[i][j]); } } return res == Integer.MAX_VALUE ? -1 : res ; } //nothing much, just bfs, but updates how many building can an empty spot reach, and //if current building can reach all other buildings, otherwise returns false, solution returns -1 private boolean bfs(int[][]grid, int i, int j, int[][] d, boolean[][] used, int count, int[][] cc) { Queue<Integer> is = new LinkedList<>(); Queue<Integer> js = new LinkedList<>(); Queue<Integer> ds = new LinkedList<>(); is.add(i); js.add(j); ds.add(0); used[i][j] = true; int[] xx = new int[]{-1,0,1,0}; int[] yy = new int[]{0,-1,0,1}; int c = 1;//used to check whether this house can reach all houses int neigh = 0; //used to check whether it can reach any empty land while(!is.isEmpty()){ int x = is.poll(); int y = js.poll(); int dis = ds.poll(); for(int k =0 ;k<4;k++){ int newx = x + xx[k]; int newy = y + yy[k]; if(newx >=0 && newx < grid.length && newy >=0 && newy<grid[0].length && !used[newx][newy]) { if(grid[newx][newy] == 0) { is.add(newx); js.add(newy); ds.add(dis+1); used[newx][newy] = true; d[newx][newy] += dis+1; cc[newx][newy]++; neigh++; } else if(grid[newx][newy] == 1) { c++; used[newx][newy] = true; } } } } return c == count && neigh > 0; }
https://leetcode.com/discuss/74441/share-my-easy-java-solution-with-comments
private class Point { public int i; public int j; public int steps; public Point(int i, int j, int steps){ this.i = i; this.j = j; this.steps = steps; } } public int shortestDistance(int[][] grid) { int height = grid.length, width = grid[0].length; int min = Integer.MAX_VALUE, houses = 0; // Count all existing houses in grid for(int i=0; i<height; i++){ for(int j=0; j<width; j++){ if(grid[i][j] == 1) houses++; } } // Start from all empty cells one by one for(int i=0; i<height; i++){ for(int j=0; j<width; j++){ if(grid[i][j] == 0){ int sum = search(i, j, houses, grid); min = Math.min(min, sum); } } } // Can't find all existing houses if(min == Integer.MAX_VALUE) return -1; else return min; } public int search(int i, int j, int houses, int[][] grid){ int height = grid.length, width = grid[0].length; boolean[][] marked = new boolean[height][width]; int houseCount = 0, sum = 0; Queue<Point> q = new LinkedList<>(); q.add(new Point(i, j, 0)); marked[i][j] = true; while(!q.isEmpty()){ Point point = q.poll(); // Traverse four surrounding cells for(int m=-1; m<=1; m++){ for(int n=-1; n<=1; n++){ // Current point itself or out of boundary if((Math.abs(m) == Math.abs(n)) || !(point.i+m >=0 && point.i+m<height && point.j+n>=0 && point.j+n<width) || marked[point.i+m][point.j+n]) continue; int cell = grid[point.i+m][point.j+n]; // Find a unmarked house if(cell == 1){ houseCount++; sum += point.steps+1; marked[point.i+m][point.j+n] = true; } else if(cell == 0){ q.add(new Point(point.i+m, point.j+n, point.steps+1)); marked[point.i+m][point.j+n] = true; } } } } // Can't find all existing houses if(houseCount < houses) return Integer.MAX_VALUE; else return sum; } }
https://evanyang.gitbooks.io/leetcode/content/LeetCode/shortest_distance_from_all_buildings.html
-- good naming
https://leetcode.com/discuss/74526/java-and-python-solutions-with-bfs
public int shortestDistance(int[][] grid) { if (grid == null || grid.length == 0 || grid[0] == null || grid[0].length == 0) return -1; List<int[]> emptyLands = new ArrayList<>(); List<int[]> buildings = new ArrayList<>(); int nrow = grid.length; int ncol = grid[0].length; int[][] dists = new int[nrow][ncol]; int[][] visitedNums = new int[nrow][ncol]; int minimalDist = Integer.MAX_VALUE; // Find Buildings and Empty Lands for (int i = 0; i < nrow; i++) { for (int j = 0; j < ncol; j++) { if (grid[i][j] == 0) { emptyLands.add(new int[]{i, j}); } else if (grid[i][j] == 1) { buildings.add(new int[]{i, j}); } } } //BFS for each Building for (int[] indices: buildings) { Queue<int[]> queue = new ArrayDeque<>(); boolean[][] visited = new boolean[nrow][ncol]; queue.offer(new int[]{indices[0], indices[1], 0}); while (!queue.isEmpty()) { int[] current = queue.poll(); int x = current[0], y = current[1], dist = current[2]; if (x+1 < nrow && grid[x+1][y] == 0 && !visited[x+1][y]) { dists[x+1][y] += dist + 1; queue.offer(new int[]{x+1, y, dist+1}); visited[x+1][y] = true; visitedNums[x+1][y]++; } if (x-1 >= 0 && grid[x-1][y] == 0 && !visited[x-1][y]) { dists[x-1][y] += dist + 1; queue.offer(new int[]{x-1, y, dist+1}); visited[x-1][y] = true; visitedNums[x-1][y]++; } if (y+1 < ncol && grid[x][y+1] == 0 && !visited[x][y+1]) { dists[x][y+1] += dist + 1; queue.offer(new int[]{x, y+1, dist+1}); visited[x][y+1] = true; visitedNums[x][y+1]++; } if (y-1 >= 0 && grid[x][y-1] == 0 && !visited[x][y-1]) { dists[x][y-1] += dist + 1; queue.offer(new int[]{x, y-1, dist+1}); visited[x][y-1] = true; visitedNums[x][y-1]++; } } } //Find the Empty Land with smallest total dist int size = buildings.size(); for (int[] indices: emptyLands) { int x = indices[0], y = indices[1]; if (dists[x][y] < minimalDist && visitedNums[x][y] == size) { minimalDist = dists[x][y]; } } return minimalDist == Integer.MAX_VALUE ? -1 : minimalDist; }

http://www.geeksforgeeks.org/number-decisions-reach-destination/
Given a grid which consists of 4 types of characters : ‘B’ ‘.’ ‘S’ and ‘D’. We need to reach D starting from S, at each step we can go to neighboring cells i.e. up, down, left and right. Cells having character ‘B’ are blocked i.e. at any step we can’t move to cell having ‘B’. Given grid has dots in such a way that there is only one way to reach any cell from any other cell. We need to tell how many times we need to choose our way from more than one choices i.e. decide the path to reach D.
We can solve this problem using DFS. In path from source to destination we can see that whenever we have more than 1 neighbors, we need to decide our path so first we do a DFS and store the path from source to the destination in terms of child-parent array and then we move from destination to source, cell by cell using parent array and at every cell where we have more than 1 neighbors we will increase our answer by 1.
void dfs(int u, vector<int> g[], int prt[], bool visit[])
{
    visit[u] = true;
    //  loop over all unvisited neighbors
    for (int i = 0; i < g[u].size(); i++)
    {
        int v = g[u][i];
        if (!visit[v])
        {
            prt[v] = u;
            dfs(v, g, prt, visit);
        }
    }
}
// method returns decision taken to reach destination
// from source
int turnsToReachDestination(string grid[], int M)
{
    int N = grid[0].length();
    //  storing direction for neighbors
    int dx[] = {-1, 0, 1, 0};
    int dy[] = {0, -1, 0, 1};
    vector<int> g[M*N];
    bool visit[M*N] = {0};
    int prt[M*N];
    int start, dest;
    /*  initialize start and dest and
        store neighbours vector g
        If cell index is (i, j), then we can convert
        it to 1D as (i*N + j)  */
    for (int i = 0; i < M; i++)
    {
        for (int j = 0; j < N; j++)
        {
            if (grid[i][j] == 'D')
                dest = i*N + j;
            if (grid[i][j] == 'S')
                start = i*N + j;
            g[i*N + j].clear();
            if (grid[i][j] != 'B')
            {
                for (int k = 0; k < 4; k++)
                {
                    int u = i + dx[k];
                    int v = j + dy[k];
                    // if neighboring cell is in boundry
                    // and doesn't have 'B'
                    if (u >= 0 && u < M && v >= 0 &&
                        v < N && grid[u][v] != 'B')
                        g[i*N + j].push_back(u*N + v);
                }
            }
        }
    }
    //  call dfs from start and fill up parent array
    dfs(start, g, part, visit);
    int curr = dest;
    int res = 0;
    //  loop from destination cell back to start cell
    while (curr != start)
    {
        /*  if current cell has more than 2 neighbors,
            then we need to decide our path to reach S
            from D, so increase result by 1 */
        if (g[curr].size() > 2)
            res++;
        curr = prt[curr];
    }
    return res;
}
http://www.cnblogs.com/icecreamdeqinw/p/5048375.html

No comments:

Post a Comment

Labels

GeeksforGeeks (1107) LeetCode (993) Algorithm (795) Review (766) to-do (633) LeetCode - Review (514) Classic Algorithm (324) Dynamic Programming (293) Classic Interview (288) Google Interview (242) Tree (145) POJ (139) Difficult Algorithm (132) LeetCode - Phone (127) EPI (125) Different Solutions (120) Bit Algorithms (118) Lintcode (113) Cracking Coding Interview (110) Smart Algorithm (109) Math (107) HackerRank (89) Binary Tree (82) Binary Search (81) Graph Algorithm (74) Greedy Algorithm (72) DFS (67) LeetCode - Extended (62) Interview Corner (61) Stack (60) List (58) Advanced Data Structure (56) BFS (54) Codility (54) ComProGuide (52) Algorithm Interview (50) Geometry Algorithm (48) Binary Search Tree (46) USACO (46) Trie (45) Mathematical Algorithm (42) ACM-ICPC (41) Interval (41) Data Structure (40) Knapsack (40) Space Optimization (40) Jobdu (39) LeetCode Hard (39) Recursive Algorithm (39) Matrix (38) String Algorithm (38) Backtracking (36) Codeforces (36) Introduction to Algorithms (36) Must Known (36) Beauty of Programming (35) Sort (35) Union-Find (34) Array (33) prismoskills (33) Segment Tree (32) Sliding Window (32) Data Structure Design (31) HDU (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) Priority Queue (27) Random (27) Graph (26) Company - LinkedIn (25) GeeksQuiz (25) Logic Thinking (25) Pre-Sort (25) hihocoder (25) Queue (24) Company-Facebook (23) High Frequency (23) Post-Order Traverse (23) TopCoder (23) Algorithm Game (22) Bisection Method (22) Hash (22) Binary Indexed Trees (21) DFS + Review (21) Lintcode - Review (21) Brain Teaser (20) CareerCup (20) Company - Twitter (20) Merge Sort (20) O(N) (20) Follow Up (19) Time Complexity (19) Two Pointers (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) Number (14) Number Theory (14) Amazon Interview (13) Basic Algorithm (13) Codechef (13) Euclidean GCD (13) KMP (13) Long Increasing Sequence(LIS) (13) Majority (13) mitbbs (13) Combination (12) Computational Geometry (12) LeetCode - Classic (12) Modify Tree (12) Reconstruct Tree (12) Reservoir Sampling (12) Reverse Thinking (12) 尺取法 (12) AOJ (11) DFS+Backtracking (11) Fast Power Algorithm (11) Graph DFS (11) LCA (11) LeetCode - DFS (11) Miscs (11) Princeton (11) Proof (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) Rolling Hash (10) SPOJ (10) Theory (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) TreeMap (9) Use XOR (9) Book Notes (8) Bottom-Up (8) Company-Amazon (8) DFS+BFS (8) LeetCode - DP (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) Interval Tree (7) Inversion (7) Kadane’s Algorithm (7) Level Order Traversal (7) Math-Divisible (7) Probability DP (7) Quick Select (7) Radix Sort (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) TreeSet (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) 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) LeetCode - TODO (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) Pascal's Triangle (3) Pattern Match (3) Project Euler (3) Rectangle (3) Scala (3) SegmentFault (3) Shuffle (3) Sieve of Eratosthenes (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) Parser (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) Skyline (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) 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) 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 - Detail (1) LeetCode - Related (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) 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