417. Pacific Atlantic Water Flow

Description

Given anm x nmatrix of non-negative integers representing the height of each unit cell in a continent, the "Pacific ocean" touches the left and top edges of the matrix and the "Atlantic ocean" touches the right and bottom edges.

Water can only flow in four directions (up, down, left, or right) from a cell to another one with height equal or lower.

Find the list of grid coordinates where water can flow to both the Pacific and Atlantic ocean.

Note:

1. The order of returned grid coordinates does not matter.
2. Both m and n are less than 150.

Example:

Given the following 5x5 matrix:

  Pacific ~   ~   ~   ~   ~ 
       ~  1   2   2   3  (5) *
       ~  3   2   3  (4) (4) *
       ~  2   4  (5)  3   1  *
       ~ (6) (7)  1   4   5  *
       ~ (5)  1   1   2   4  *
          *   *   *   *   * Atlantic

Return:

[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (positions with parentheses in above matrix).

Discussion

Solution 1

dfs, tracker[m][n], visited_pacific[m][n], visited_atlantic[m][n].

for every matrix[y][x] >= matrix[preY][preX], tracker[y][x] -= 1;

dfs from out-most points.

if tracker[y][x] == -2, it is the candidate.

class Solution {
public:
    vector<pair<int, int>> pacificAtlantic(vector<vector<int>>& matrix) {
        vector<pair<int, int>> rst;
        if (0 == matrix.size()) 
            return rst;

        vector<vector<int>> tracker(matrix.size(), vector<int>(matrix[0].size(), 0));
        vector<vector<bool>> visted_pacific(matrix.size(), vector<bool>(matrix[0].size(), false));
        vector<vector<bool>> visted_atlantic(matrix.size(), vector<bool>(matrix[0].size(), false));

        for(int y = 0; y < matrix.size(); ++y) {
            dfs(tracker, matrix, visted_pacific, y, 0);
            dfs(tracker, matrix, visted_atlantic, y, matrix[0].size() - 1);
        }

        /*
        for(int y = 0; y < matrix.size(); ++y) {
            for(int x = 0; x < matrix[0].size(); ++x) {
                visted_pacific[y][x] = false;
                visted_atlantic[y][x] = false;
            }
        }
        */

        for(int x = 0; x < matrix[0].size(); ++x) {
            dfs(tracker, matrix, visted_pacific, 0, x);
            dfs(tracker, matrix, visted_atlantic, matrix.size() - 1, x);
        }

        #ifdef TEST
        for(int y= 0; y < matrix.size(); ++y) {
            copy(tracker[y].begin(), tracker[y].end(), ostream_iterator<int>(cout, ", "));
            cout << endl;
        }
        cout << "test" << endl;
        #endif


        for(int y = 0; y < tracker.size(); ++y) {
            for(int x = 0; x < tracker[0].size(); ++x) {
                if(-2 == tracker[y][x]) {
                    rst.emplace_back(make_pair(y, x));
                }
                //cout << y << "," << x << endl;
            }
        }
        //cout << "test" << endl;

        #ifdef TEST
            for(auto each : rst) {
                cout << "[" << each.first << ", " << each.second << "]" << endl;
            }
        #endif
        return rst;
    }
private:
    void dfs(vector<vector<int>>& tracker, vector<vector<int>>& matrix, 
             vector<vector<bool>>& visited, 
             int y, int x) {
        vector<int> dy = {0,0,1,-1};
        vector<int> dx = {1,-1,0,0};

        if(visited[y][x] == true)
            return;

        tracker[y][x] -= 1;
        visited[y][x] = true;

        for(int loop = 0; loop < dy.size(); ++loop) {
            if(y + dy[loop] >= 0 && y + dy[loop] < tracker.size() &&
               x + dx[loop] >= 0 && x + dx[loop] < tracker[0].size() && 
               visited[y+dy[loop]][x+dx[loop]] == false && 
               matrix[y+dy[loop]][x+dx[loop]] >= matrix[y][x]) {
                dfs(tracker, matrix, visited, y + dy[loop], x + dx[loop]);
            }
        }
    }
};