<p>A robot on an infinite XY-plane starts at point <code>(0, 0)</code> facing north. The robot receives an array of integers <code>commands</code>, which represents a sequence of moves that it needs to execute. There are only three possible types of instructions the robot can receive:</p>

<ul>
	<li><code>-2</code>: Turn left <code>90</code> degrees.</li>
	<li><code>-1</code>: Turn right <code>90</code> degrees.</li>
	<li><code>1 &lt;= k &lt;= 9</code>: Move forward <code>k</code> units, one unit at a time.</li>
</ul>

<p>Some of the grid squares are <code>obstacles</code>. The <code>i<sup>th</sup></code> obstacle is at grid point <code>obstacles[i] = (x<sub>i</sub>, y<sub>i</sub>)</code>. If the robot runs into an obstacle, it will stay in its current location (on the block adjacent to the obstacle) and move onto the next command.</p>

<p>Return the <strong>maximum squared Euclidean distance</strong> that the robot reaches at any point in its path (i.e. if the distance is <code>5</code>, return <code>25</code>).</p>

<p><strong>Note:</strong></p>

<ul>
	<li>There can be an obstacle at <code>(0, 0)</code>. If this happens, the robot will ignore the obstacle until it has moved off the origin. However, it will be unable to return to <code>(0, 0)</code> due to the obstacle.</li>
	<li>North means +Y direction.</li>
	<li>East means +X direction.</li>
	<li>South means -Y direction.</li>
	<li>West means -X direction.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">commands = [4,-1,3], obstacles = []</span></p>

<p><strong>Output:</strong> <span class="example-io">25</span></p>

<p><strong>Explanation: </strong></p>

<p>The robot starts at <code>(0, 0)</code>:</p>

<ol>
	<li>Move north 4 units to <code>(0, 4)</code>.</li>
	<li>Turn right.</li>
	<li>Move east 3 units to <code>(3, 4)</code>.</li>
</ol>

<p>The furthest point the robot ever gets from the origin is <code>(3, 4)</code>, which squared is <code>3<sup>2</sup> + 4<sup>2 </sup>= 25</code> units away.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">commands = [4,-1,4,-2,4], obstacles = [[2,4]]</span></p>

<p><strong>Output:</strong> <span class="example-io">65</span></p>

<p><strong>Explanation:</strong></p>

<p>The robot starts at <code>(0, 0)</code>:</p>

<ol>
	<li>Move north 4 units to <code>(0, 4)</code>.</li>
	<li>Turn right.</li>
	<li>Move east 1 unit and get blocked by the obstacle at <code>(2, 4)</code>, robot is at <code>(1, 4)</code>.</li>
	<li>Turn left.</li>
	<li>Move north 4 units to <code>(1, 8)</code>.</li>
</ol>

<p>The furthest point the robot ever gets from the origin is <code>(1, 8)</code>, which squared is <code>1<sup>2</sup> + 8<sup>2</sup> = 65</code> units away.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">commands = [6,-1,-1,6], obstacles = [[0,0]]</span></p>

<p><strong>Output:</strong> <span class="example-io">36</span></p>

<p><strong>Explanation:</strong></p>

<p>The robot starts at <code>(0, 0)</code>:</p>

<ol>
	<li>Move north 6 units to <code>(0, 6)</code>.</li>
	<li>Turn right.</li>
	<li>Turn right.</li>
	<li>Move south 5 units and get blocked by the obstacle at <code>(0,0)</code>, robot is at <code>(0, 1)</code>.</li>
</ol>

<p>The furthest point the robot ever gets from the origin is <code>(0, 6)</code>, which squared is <code>6<sup>2</sup> = 36</code> units away.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= commands.length &lt;= 10<sup>4</sup></code></li>
	<li><code>commands[i]</code> is either <code>-2</code>, <code>-1</code>, or an integer in the range <code>[1, 9]</code>.</li>
	<li><code>0 &lt;= obstacles.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-3 * 10<sup>4</sup> &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 3 * 10<sup>4</sup></code></li>
	<li>The answer is guaranteed to be less than <code>2<sup>31</sup></code>.</li>
</ul>

class Solution {
public:
    int robotSim(vector<int>& commands, vector<vector<int>>& obstacles) {
        ios::sync_with_stdio(false);
        cin.tie(0);
        cout.tie(0);
        unordered_map<int,set<int>> obst_x;
        unordered_map<int,set<int>> obst_y;
        int ans = INT_MIN;
        for(int i=0;i<obstacles.size();i++){
            int x = obstacles[i][0];
            int y = obstacles[i][1];
            obst_x[x].insert(y);
            obst_y[y].insert(x);
        }
        vector<int> dy = {1,0,-1,0};
        vector<int> dx = {0,1,0,-1};
        pair<int,int> p = {0,0};
        int d = 0;
        for(int i=0;i<commands.size();i++){
            if (commands[i] ==-1){
                if(d==3) d = 0;
                else d++;

            }else if (commands[i]==-2){
                if(d==0) d = 3;
                else d--;
                
            }else{
                int a = commands[i];
                cout<<a<<endl;
                int x = dx[d]*a;
                int y = dy[d]*a;
                int oldx = p.first;
                int oldy = p.second;
                int newx = p.first+x;
                int newy = p.second+y;
                p = {newx,newy};
                if (x==0){
                    if (y>0){
                    for(int j=oldy+1;j<=newy;j++){
                        if(obst_x[oldx].count(j)==1){
                            p = {oldx,j-1};
                        }
                    }
                    }else{
                        for(int j=oldy-1;j>=newy;j--){
                        if(obst_x[oldx].count(j)==1){
                            p = {oldx,j+1};
                        }
                    }   
                    }
                }else if (y==0){
                    if(x>0){
                    for(int j = oldx+1;j<=newx;j++){
                        if (obst_y[oldy].count(j)==1){
                            p = {j-1,oldy};
                        }
                    }
                    }else{
                        for(int j = oldx-1;j>=newx;j--){
                        if (obst_y[oldy].count(j)==1){
                            p = {j+1,oldy};
                        }
                    }
                    }
                }
                cout<<p.first<<" "<<p.second<<endl;
                int h = pow(p.first,2)+pow(p.second,2);
                ans = max(h,ans);
            }

        }
        return ans;
    }
};
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