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Author: koralkulacoglu

LeetCode/3338-count-submatrices-with-top-left-element-and-sum-less-than-k/README.md

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+<p>You are given a <strong>0-indexed</strong> integer matrix <code>grid</code> and an integer <code>k</code>.</p>
+
+<p>Return <em>the <strong>number</strong> of <span data-keyword="submatrix">submatrices</span> that contain the top-left element of the</em> <code>grid</code>, <em>and have a sum less than or equal to </em><code>k</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2024/01/01/example1.png" style="padding: 10px; background: #fff; border-radius: .5rem;" />
+<pre>
+<strong>Input:</strong> grid = [[7,6,3],[6,6,1]], k = 18
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> There are only 4 submatrices, shown in the image above, that contain the top-left element of grid, and have a sum less than or equal to 18.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2024/01/01/example21.png" style="padding: 10px; background: #fff; border-radius: .5rem;" />
+<pre>
+<strong>Input:</strong> grid = [[7,2,9],[1,5,0],[2,6,6]], k = 20
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> There are only 6 submatrices, shown in the image above, that contain the top-left element of grid, and have a sum less than or equal to 20.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>m == grid.length </code></li>
+	<li><code>n == grid[i].length</code></li>
+	<li><code>1 &lt;= n, m &lt;= 1000 </code></li>
+	<li><code>0 &lt;= grid[i][j] &lt;= 1000</code></li>
+	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
+</ul>

LeetCode/3338-count-submatrices-with-top-left-element-and-sum-less-than-k/solution.cpp

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+class Solution {
+public:
+    int countSubmatrices(vector<vector<int>>& grid, int k) {
+        int n = grid.size();
+        int m = grid[0].size();
+        int ans = 0;
+        for (int i=0; i<n; i++) {
+            for (int j=0; j<m; j++) {
+                if (i > 0) grid[i][j] += grid[i-1][j];
+                if (j > 0) grid[i][j] += grid[i][j-1];
+                if (i > 0 && j > 0) grid[i][j] -= grid[i-1][j-1];
+                ans += grid[i][j] <= k;
+            }
+        }
+        return ans;
+    }
+};