Sync LeetCode submission Runtime - 24 ms (88.63%), Memory - 18.4 MB (67.37%)

Author: jimit105

1049-minimum-domino-rotations-for-equal-row/README.md

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+<p>In a row of dominoes, <code>tops[i]</code> and <code>bottoms[i]</code> represent the top and bottom halves of the <code>i<sup>th</sup></code> domino. (A domino is a tile with two numbers from 1 to 6 - one on each half of the tile.)</p>
+
+<p>We may rotate the <code>i<sup>th</sup></code> domino, so that <code>tops[i]</code> and <code>bottoms[i]</code> swap values.</p>
+
+<p>Return the minimum number of rotations so that all the values in <code>tops</code> are the same, or all the values in <code>bottoms</code> are the same.</p>
+
+<p>If it cannot be done, return <code>-1</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/05/14/domino.png" style="height: 300px; width: 421px;" />
+<pre>
+<strong>Input:</strong> tops = [2,1,2,4,2,2], bottoms = [5,2,6,2,3,2]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> 
+The first figure represents the dominoes as given by tops and bottoms: before we do any rotations.
+If we rotate the second and fourth dominoes, we can make every value in the top row equal to 2, as indicated by the second figure.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> tops = [3,5,1,2,3], bottoms = [3,6,3,3,4]
+<strong>Output:</strong> -1
+<strong>Explanation:</strong> 
+In this case, it is not possible to rotate the dominoes to make one row of values equal.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= tops.length &lt;= 2 * 10<sup>4</sup></code></li>
+	<li><code>bottoms.length == tops.length</code></li>
+	<li><code>1 &lt;= tops[i], bottoms[i] &lt;= 6</code></li>
+</ul>

1049-minimum-domino-rotations-for-equal-row/solution.py

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+# Approach: Greedy
+
+# Time: O(n)
+# Space: O(1)
+
+class Solution:
+    def minDominoRotations(self, tops: List[int], bottoms: List[int]) -> int:
+        n = len(tops)
+
+        def check(x):
+            # Count rotations needed to make all tops x
+            rotations_top = 0
+            
+            # Count rotations needed to make all bottoms x
+            rotations_bottom = 0
+
+            for i in range(n):
+                # If neither top nor bottom is x, impossible
+                if tops[i] != x and bottoms[i] != x:
+                    return float('inf')
+                # If top is not x but bottom is x
+                elif tops[i] != x:
+                    rotations_top += 1
+                # If bottom is not x but top is x
+                elif bottoms[i] != x:
+                    rotations_bottom += 1
+
+            return min(rotations_top, rotations_bottom)
+
+        # Try making everything the value of first domino's bottom or first domino's bottom
+        min_rotations = min(check(tops[0]), check(bottoms[0]))
+
+        return -1 if min_rotations == float('inf') else min_rotations