Sync LeetCode submission Runtime - 155 ms (19.74%), Memory - 43.1 MB (77.33%)

Author: jimit105

2689-rearranging-fruits/README.md

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+<p>You have two fruit baskets containing <code>n</code> fruits each. You are given two <strong>0-indexed</strong> integer arrays <code>basket1</code> and <code>basket2</code> representing the cost of fruit in each basket. You want to make both baskets <strong>equal</strong>. To do so, you can use the following operation as many times as you want:</p>
+
+<ul>
+	<li>Choose two indices <code>i</code> and <code>j</code>, and swap the <code>i<sup><font size="1">th</font></sup></code> fruit of <code>basket1</code> with the <code>j<sup><font size="1">th</font></sup></code> fruit of <code>basket2</code>.</li>
+	<li>The cost of the swap is <code>min(basket1[i], basket2[j])</code>.</li>
+</ul>
+
+<p>Two baskets are considered equal if sorting them according to the fruit cost makes them exactly the same baskets.</p>
+
+<p>Return <em>the minimum cost to make both the baskets equal or </em><code>-1</code><em> if impossible.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> basket1 = [4,2,2,2], basket2 = [1,4,1,2]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> Swap index 1 of basket1 with index 0 of basket2, which has cost 1. Now basket1 = [4,1,2,2] and basket2 = [2,4,1,2]. Rearranging both the arrays makes them equal.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> basket1 = [2,3,4,1], basket2 = [3,2,5,1]
+<strong>Output:</strong> -1
+<strong>Explanation:</strong> It can be shown that it is impossible to make both the baskets equal.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>basket1.length == basket2.length</code></li>
+	<li><code>1 &lt;= basket1.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= basket1[i], basket2[i] &lt;= 10<sup>9</sup></code></li>
+</ul>

2689-rearranging-fruits/solution.py

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+# Approach: Greedy
+
+# Time: O(n log n)
+# Space: O(n)
+
+from collections import Counter
+
+class Solution:
+    def minCost(self, basket1: List[int], basket2: List[int]) -> int:
+        freq = Counter()
+        m = float('inf')
+        
+        for b1 in basket1:
+            freq[b1] += 1
+            m = min(m, b1)
+        for b2 in basket2:
+            freq[b2] -= 1
+            m = min(m, b2)
+
+        merge = []
+        for k, c in freq.items():
+            if c % 2 != 0:
+                return -1
+            merge.extend([k] * (abs(c) // 2))
+
+        if not merge:
+            return 0
+        merge.sort()
+        return sum(min(2 * m, x) for x in merge[: len(merge) // 2])
+