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Commit 6ac5d90

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Sync LeetCode submission Runtime - 71 ms (83.57%), Memory - 34.1 MB (28.45%)
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<p>You are given a <strong>0-indexed</strong> integer array <code>players</code>, where <code>players[i]</code> represents the <strong>ability</strong> of the <code>i<sup>th</sup></code> player. You are also given a <strong>0-indexed</strong> integer array <code>trainers</code>, where <code>trainers[j]</code> represents the <strong>training capacity </strong>of the <code>j<sup>th</sup></code> trainer.</p>
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<p>The <code>i<sup>th</sup></code> player can <strong>match</strong> with the <code>j<sup>th</sup></code> trainer if the player&#39;s ability is <strong>less than or equal to</strong> the trainer&#39;s training capacity. Additionally, the <code>i<sup>th</sup></code> player can be matched with at most one trainer, and the <code>j<sup>th</sup></code> trainer can be matched with at most one player.</p>
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<p>Return <em>the <strong>maximum</strong> number of matchings between </em><code>players</code><em> and </em><code>trainers</code><em> that satisfy these conditions.</em></p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> players = [4,7,9], trainers = [8,2,5,8]
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<strong>Output:</strong> 2
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<strong>Explanation:</strong>
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One of the ways we can form two matchings is as follows:
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- players[0] can be matched with trainers[0] since 4 &lt;= 8.
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- players[1] can be matched with trainers[3] since 7 &lt;= 8.
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It can be proven that 2 is the maximum number of matchings that can be formed.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> players = [1,1,1], trainers = [10]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong>
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The trainer can be matched with any of the 3 players.
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Each player can only be matched with one trainer, so the maximum answer is 1.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= players.length, trainers.length &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= players[i], trainers[j] &lt;= 10<sup>9</sup></code></li>
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</ul>
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<p>&nbsp;</p>
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<p><strong>Note:</strong> This question is the same as <a href="https://leetcode.com/problems/assign-cookies/description/" target="_blank"> 445: Assign Cookies.</a></p>
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# Approach: Sorting + Two Pointers + Greedy
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# m = len(players), n = len(trainers)
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# Time: O(m log m + n log n)
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# Space: O(log m + log n)
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class Solution:
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def matchPlayersAndTrainers(self, players: List[int], trainers: List[int]) -> int:
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players.sort()
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trainers.sort()
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m, n = len(players), len(trainers)
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i = j = count = 0
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while i < m and j < n:
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while j < n and players[i] > trainers[j]:
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j += 1
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if j < n:
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count += 1
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i += 1
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j += 1
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return count
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