As Far From Land As Possible in C++

Suppose we have one N x N grid containing only values like 0 and 1, where 0 represents water and 1 represents the land, we have to find a water cell such that its distance to the nearest land cell is maximized and return the distance. Here we will use the Manhattan distance − the distance between two cells (x0, y0) and (x1, y1) is |x0 - x1| + |y0 - y1|. If no land or water is present in the grid, then return -1.

Then the output will be 2, as the cell (1,1) is as far as possible from all the land with distance 2.

To solve this, we will follow these steps −

• dir := [(1, 0), (-1, 0), (1, -1), (1, 1), (-1, 1), (-1, -1), (0, 1), (0, -1)]

• dir2 := [(1, 0), (-1, 0), (0, 1), (0, -1)]

• Define a map m. Define a queue q. n := row count and c := column count

• for i in range 0 to n – 1

  • for j in range 0 to n – 1

    • if grid[i, j] is 1, then insert a pair (i, j) into q and put m[(i, j)] := (j ,i)

• ret := -1

• while the q is not empty

  • sz := size of q

  • while sz is not 0

    • temp := first element of q, delete first element from q

    • for k in range 0 to 3 −

      • nx := first value of temp + dir2[k, 0]

      • ny := second value of temp + dir2[k, 1]

      • if nx and ny are not in range of grid, or grid[nx, ny] is 1, then skip to the next iteration.

      • m[(nx, ny)] := m[temp]

      • ret := max of (distance of (nx, ny) and m(temp)) and ret

      • insert (nx,ny) into q

      • set grid[nx, ny] := 1

    • decrease sz by 1

• return ret

Example(C++)

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
int dir[8][2] = {
   {1, 0}, {-1, 0}, {1, -1}, {1, 1},
   {-1, 1}, {-1, -1}, {0, 1}, {0, -1}
};
int dir2[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
class Solution {
   public:
   int calcDist(int x1, int y1, int x2, int y2){
      return abs(x1 - x2) + abs(y1 - y2);
   }
   int maxDistance(vector<vector<int>>& grid) {
      map < pair <int, int>, pair <int, int> > m;
      queue < pair <int, int> > q;
      int n = grid.size();
      int c = n? grid[0].size() : 0;
      for(int i = 0; i < n; i++){
         for(int j = 0; j < c; j++){
            if(grid[i][j] == 1){
               q.push({i, j});
               m[{i, j}] = {i, j};
            }
         }
      }
      int ret = -1;
      while(!q.empty()){
         int sz = q.size();
         while(sz--){
            pair <int, int> temp = q.front();
            q.pop();
            for(int k = 0; k < 4; k++){
               int nx = temp.first + dir2[k][0];
               int ny = temp.second + dir2[k][1];
               if(nx < 0 || ny < 0 || nx >= n || ny >= c || grid[nx][ny]) continue;
               m[{nx, ny}] = m[temp];
               ret = max(calcDist(nx, ny, m[temp].first,
               m[temp].second), ret);
               q.push({nx, ny});
               grid[nx][ny] = 1;
            }
         }
      }
      return ret;
   }
};
main(){
   vector<vector<int>> v1 = {{1,0,1},{0,0,0},{1,0,1}};
   Solution ob;
   cout << (ob.maxDistance(v1));
}

Input

["alice,20,800,mtv","bob,50,1200,mtv"]

Output

2