Minimum obstacle removal to reach corner

//BFS : dijkstra's class Solution { public int minimumObstacles(int[][] grid) { int minObstacleRemoved = Integer.MAX_VALUE; Queue<Cell> q = new PriorityQueue<>((a,b)-> Integer.compare(a.c,b.c)); int visited[][] = new int[grid.length][grid[0].length]; q.add(new Cell(0,0,0)); while(!q.isEmpty()){ Cell cell = q.remove(); int I = cell.i; int J = cell.j; if(visited[I][J] ==1) continue; visited[I][J] = 1; if(I == grid.length-1 && J == grid[0].length-1) { minObstacleRemoved = cell.c; break; } int dirs[][] = {{0,-1},{0,1},{-1,0},{1,0}}; int obstacleRemovedSoFar = cell.c; for(int dir[]: dirs){ int i = I + dir[0]; int j = J + dir[1]; if(i>=0 && j>=0 && i<grid.length && j< grid[0].length && visited[i][j]==0){ int c = obstacleRemovedSoFar; //if it is an obstacle then 1 more obstacle you will have to remove if(grid[i][j]==1) c+=1; q.add(new Cell(i,j,c)); } } } return minObstacleRemoved; } } class Cell{ int i; int j; int c; public Cell(int i, int j, int c){ this.i = i; this.j = j; this.c = c; } }