java - Find available "number" in a 2d array -

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i have problem need solve in effecient way. have 2d array contains following: 1 "wall" means cannot go through it. 2 entrance "enter" array or map if like. 3 things need find. here example of map:

1111111 1  3131 2 11111 1    31 1111111

this example of array need in. can see there 3 "unreachable, since it's surrounded wall "1". means there 2 available numbers in array.

first need find entrance. since entrance can anywhere need search entire array. have done following:

int treasureamount = 0;      point entrance = new point(0,0);      (int = 0; < n; i++) {          (int j = 0; j < n; i++){              if(map[i][j] == 2){                  entrance.x =i;                  entrance.y =j;              }           }

this takes o(n^2) time, , don't see way this, since entrance can anywhere. i'm not sure how find available numbers effectivly , fast. thought while searching arrays entrance @ same time find number 3 in array though might not accessible, , after i'm not sure how effectivly find accessible.

you cannot better o(n^2). take time read array. depth first search find reachable 3's in array. here pseudo code.

main() {     read array , mark entrance ent.x , ent.y , array threex[] , threey[] stores exit position.     boolean visited[][]; //stores whether array[i][j] reachable or not.     dfs(ent.x,ent.y);     each element in 3 arrays     {         if(visited[threex[i]][threey[i]]) print ("reachable");         else print("not reachable", threex[i], threey[i]);     } } int dx[]={1,0,-1,0},dy[]={0,1,0,-1}; // dx[i], dy[i] tells whether move in e,n,w,s respectively. dfs(int x,int y) {     visited[x][y]=true;     for(i=0;i<4;i++)//move in directions     {         int newx=x+dx[i],newy=y+dy[i];         //check if within array boundary         if(newx>=0&&newx<n && newy>=0&&newy<n)         if(!visited[newx][newy] && array[newx][newy]!=1) // check if node unvisited , pemissible              dfs(newx,newy);     } }

since each array element taken not more once in dfs function complexity of code o(n^2).


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