Walls And Gates | Graph traversal problem solution

 Walls And Gates | Leetcode 286 solution 

You are given a m x n 2D grid initialized with these three possible values.

1. -1 - A wall or an obstacle.
2. 0 - A gate.
3. INF - Infinity means an empty room. We use the value 231 - 1 = 2147483647 to represent INF as you 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.

Example: 

Given the 2D grid:

INF  -1  0  INF
INF INF INF  -1
INF  -1 INF  -1
  0  -1 INF INF

After running your function, the 2D grid should be:

  3  -1   0   1
  2   2   1  -1
  1  -1   2  -1
  0  -1   3   4

Solution : 

#include <bits/stdc++.h> vector<vector<int>> wallsAndGates(vector<vector<int>> &a, int n, int m) {     // Write your code here.     queue<pair<int , int>>q;     // vector<pair<int, int>>vect;                   vector<vector<int>>visited(n , vector<int>(m , 0));     for(int i =0; i < n ; i++ )     {         for(int j =0; j < m ; j++)         {                         if(a[i][j] == 0)             {                 visited[i][j] =1;                 q.push({i, j});             }         }     }             //    visited[vect[l].first][vect[l].second] =1;     while(!q.empty())     {         auto node = q.front();         int i = node.first;         int j = node.second;         q.pop();         int x[4] = {0 , 1 , 0, -1};         int y[4] = {1 , 0 , -1 , 0};         for(int k =0 ; k < 4 ; k++)         {             int row = i + x[k];             int col = j + y[k];             if(row >=0  && row < n && col >=0 && col < m && !visited[row][col] &&             a[row][col] != -1 && a[row][col] != 0)             {                 visited[row][col] = 1;                 a[row][col] = min(a[row][col], a[i][j] +1 );                 q.push({row, col});             }         }     }     return a; }