Range Sum Query 2D - Immutable
<p>Given a 2D matrix <code>matrix</code>, handle multiple queries of the following type:</p>

<ul>
	<li>Calculate the <strong>sum</strong> of the elements of <code>matrix</code> inside the rectangle defined by its <strong>upper left corner</strong> <code>(row1, col1)</code> and <strong>lower right corner</strong> <code>(row2, col2)</code>.</li>
</ul>

<p>Implement the <code>NumMatrix</code> class:</p>

<ul>
	<li><code>NumMatrix(int[][] matrix)</code> Initializes the object with the integer matrix <code>matrix</code>.</li>
	<li><code>int sumRegion(int row1, int col1, int row2, int col2)</code> Returns the <strong>sum</strong> of the elements of <code>matrix</code> inside the rectangle defined by its <strong>upper left corner</strong> <code>(row1, col1)</code> and <strong>lower right corner</strong> <code>(row2, col2)</code>.</li>
</ul>

<p>You must design an algorithm where <code>sumRegion</code> works on <code>O(1)</code> time complexity.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0300-0399/0304.Range%20Sum%20Query%202D%20-%20Immutable/images/sum-grid.jpg" style="width: 415px; height: 415px;" />
<pre>
<strong>Input</strong>
[&quot;NumMatrix&quot;, &quot;sumRegion&quot;, &quot;sumRegion&quot;, &quot;sumRegion&quot;]
[[[[3, 0, 1, 4, 2], [5, 6, 3, 2, 1], [1, 2, 0, 1, 5], [4, 1, 0, 1, 7], [1, 0, 3, 0, 5]]], [2, 1, 4, 3], [1, 1, 2, 2], [1, 2, 2, 4]]
<strong>Output</strong>
[null, 8, 11, 12]

<strong>Explanation</strong>
NumMatrix numMatrix = new NumMatrix([[3, 0, 1, 4, 2], [5, 6, 3, 2, 1], [1, 2, 0, 1, 5], [4, 1, 0, 1, 7], [1, 0, 3, 0, 5]]);
numMatrix.sumRegion(2, 1, 4, 3); // return 8 (i.e sum of the red rectangle)
numMatrix.sumRegion(1, 1, 2, 2); // return 11 (i.e sum of the green rectangle)
numMatrix.sumRegion(1, 2, 2, 4); // return 12 (i.e sum of the blue rectangle)

</pre>

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

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 200</code></li>
	<li><code>-10<sup>4</sup> &lt;= matrix[i][j] &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= row1 &lt;= row2 &lt; m</code></li>
	<li><code>0 &lt;= col1 &lt;= col2 &lt; n</code></li>
	<li>At most <code>10<sup>4</sup></code> calls will be made to <code>sumRegion</code>.</li>
</ul>
Design; Array; Matrix; Prefix Sum
class NumMatrix {
public:
    vector<vector<int>> s;

    NumMatrix(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        s.resize(m + 1, vector<int>(n + 1));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                s[i + 1][j + 1] = s[i + 1][j] + s[i][j + 1] - s[i][j] + matrix[i][j];
            }
        }
    }

    int sumRegion(int row1, int col1, int row2, int col2) {
        return s[row2 + 1][col2 + 1] - s[row2 + 1][col1] - s[row1][col2 + 1] + s[row1][col1];
    }
};

/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix* obj = new NumMatrix(matrix);
 * int param_1 = obj->sumRegion(row1,col1,row2,col2);
 */