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Author: jimit105

0073-set-matrix-zeroes/README.md

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+<p>Given an <code>m x n</code> integer matrix <code>matrix</code>, if an element is <code>0</code>, set its entire row and column to <code>0</code>&#39;s.</p>
+
+<p>You must do it <a href="https://en.wikipedia.org/wiki/In-place_algorithm" target="_blank">in place</a>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/08/17/mat1.jpg" style="width: 450px; height: 169px;" />
+<pre>
+<strong>Input:</strong> matrix = [[1,1,1],[1,0,1],[1,1,1]]
+<strong>Output:</strong> [[1,0,1],[0,0,0],[1,0,1]]
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/08/17/mat2.jpg" style="width: 450px; height: 137px;" />
+<pre>
+<strong>Input:</strong> matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
+<strong>Output:</strong> [[0,0,0,0],[0,4,5,0],[0,3,1,0]]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>m == matrix.length</code></li>
+	<li><code>n == matrix[0].length</code></li>
+	<li><code>1 &lt;= m, n &lt;= 200</code></li>
+	<li><code>-2<sup>31</sup> &lt;= matrix[i][j] &lt;= 2<sup>31</sup> - 1</code></li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong>Follow up:</strong></p>
+
+<ul>
+	<li>A straightforward solution using <code>O(mn)</code> space is probably a bad idea.</li>
+	<li>A simple improvement uses <code>O(m + n)</code> space, but still not the best solution.</li>
+	<li>Could you devise a constant space solution?</li>
+</ul>

0073-set-matrix-zeroes/solution.py

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+# Approach 2: Use first row and column
+
+class Solution(object):
+    def setZeroes(self, matrix: List[List[int]]) -> None:
+        is_col = False
+        R = len(matrix)
+        C = len(matrix[0])
+        for i in range(R):
+            # Since first cell for both first row and first column is the same i.e. matrix[0][0]
+            # We can use an additional variable for either the first row/column.
+            # For this solution we are using an additional variable for the first column
+            # and using matrix[0][0] for the first row.
+            if matrix[i][0] == 0:
+                is_col = True
+            for j in range(1, C):
+                # If an element is zero, we set the first element of the corresponding row and column to 0
+                if matrix[i][j] == 0:
+                    matrix[0][j] = 0
+                    matrix[i][0] = 0
+
+        # Iterate over the array once again and using the first row and first column, update the elements.
+        for i in range(1, R):
+            for j in range(1, C):
+                if not matrix[i][0] or not matrix[0][j]:
+                    matrix[i][j] = 0
+
+        # See if the first row needs to be set to zero as well
+        if matrix[0][0] == 0:
+            for j in range(C):
+                matrix[0][j] = 0
+
+        # See if the first column needs to be set to zero as well
+        if is_col:
+            for i in range(R):
+                matrix[i][0] = 0
+