<p>Given a positive integer <code>n</code>, generate an <code>n x n</code> <code>matrix</code> filled with elements from <code>1</code> to <code>n<sup>2</sup></code> in spiral order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/13/spiraln.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> [[1,2,3],[8,9,4],[7,6,5]]
</pre>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> [[1]]
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 20</code></li>
</ul>

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {

        vector<vector<int>> ans(n,vector<int>(n,0));

        int left=0;
        int right=n-1;
        int top=0;
        int bottom=n-1;

        int count=1;

        while(top<=bottom && left<=right){
            for(int row=left;row<=right;row++){
                ans[top][row]=count;
                count++;
            }
            top++;

            for(int col=top;col<=bottom;col++){
                ans[col][right]=count;
                count++;
            }
            right--;


            if(top<=bottom){
                for(int row=right;row>=left;row--){
                    ans[bottom][row]=count;
                    count++;
                }
                bottom--;
            }
            
            if(left<=right){
                for(int col=bottom;col>=top;col--){
                    ans[col][left]=count;
                    count++;
                }
                left++;
            }
            
        }

        return ans;
    }
};


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