Walls and Gates in C++

C++Server Side ProgrammingProgramming

Suppose we have one m x n 2D grid, and that is initialized with these three possible values.

  • -1 for a wall or an obstacle.

  • 0 for a gate.

  • INF This is infinity means an empty room.

Here 2^31 - 1 = 2147483647 is INF as we may assume that the distance to a gate is less than 2147483647. Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF.

So, if the input is like

INF-10INF
INFINFINF-1
INF-1INF-1
0-1INFINF

then the output will be

3-101
221-1
1-12-1
0-134

To solve this, we will follow these steps −

  • Define an array dir of size: 4 x 2 := {{1, 0}, {-1, 0}, {0, 1}, {0,-1}}

  • n := size of rooms

  • m := (if n is non-zero, then column count, otherwise 0)

  • Define one queue q of pairs

  • for initialize i := 0, when i < n, update (increase i by 1), do −

    • for initialize j := 0, when j < m, update (increase j by 1), do −

      • if rooms[i, j] is same as 0, then −

        • insert { i, j } into q

  • for initialize lvl := 1, when not q is empty, update (increase lvl by 1), do −

    • sz := size of q

    • while sz is non-zero, decrease sz by 1 in each iteration, do −

      • Define one pair curr := first element of q

      • delete element from q

      • x := curr.first

      • y := curr.second

      • for initialize i := 0, when i < 4, update (increase i by 1), do −

        • nx := x + dir[i, 0]

        • ny := y + dir[i, 1]

        • if nx < 0 or ny < 0 or nx >= n or ny >= m or rooms[nx, ny] < lvl, then −

          • Ignore following part, skip to the next iteration

        • rooms[nx, ny] := lvl

        • insert { nx, ny } into q

Example 

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<vector<auto< > v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << "[";
      for(int j = 0; j <v[i].size(); j++){
         cout << v[i][j] << ", ";
      }
      cout << "],";
   }
   cout << "]"<<endl;
}
int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
class Solution {
public:
   void wallsAndGates(vector<vector<int<>& rooms) {
      int n = rooms.size();
      int m = n ? rooms[0].size() : 0;
      queue<pair<int, int> > q;
      for (int i = 0; i < n; i++) {
         for (int j = 0; j < m; j++) {
            if (rooms[i][j] == 0)
               q.push({ i, j });
         }
      }
      for (int lvl = 1; !q.empty(); lvl++) {
         int sz = q.size();
         while (sz--) {
            pair<int, int> curr = q.front();
            q.pop();
            int x = curr.first;
            int y = curr.second;
            for (int i = 0; i < 4; i++) {
               int nx = x + dir[i][0];
               int ny = y + dir[i][1];
               if (nx < 0 || ny < 0 || nx >= n || ny >= m || rooms[nx][ny] < lvl)
                  continue;
               rooms[nx][ny] = lvl;
               q.push({ nx, ny });
            }
         }
      }
   }
};
main(){
   vector<vector<int<> v = {{2147483647,-1,0,2147483647}, {2147483647,2147483647,2147483647,-1}, {2147483647,-1,2147483647,-1}, {0,-1,2147483647,2147483647}};
   Solution ob;
   ob.wallsAndGates(v);
   print_vector(v);
}

Input

{{2147483647,-1,0,2147483647},{2147483647,2147483647,2147483647,-1}, {2147483647,-1,2147483647,-1},{0,-1,2147483647,2147483647}}

Output

[[3, -1, 0, 1, ],[2, 2, 1, -1, ],[1, -1, 2, -1, ],[0, -1, 3, 4, ],]
raja
Published on 18-Nov-2020 16:26:19
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