Maximum Matching of Players With Trainers - Solution & Explanation
Problem Statement
You are given a 0-indexed integer array players, where players[i] represents the ability of the ith player. You are also given a 0-indexed integer array trainers, where trainers[j] represents the training capacity of the jth trainer.
The ith player can match with the jth trainer if the player's ability is less than or equal to the trainer's training capacity. Additionally, the ith player can be matched with at most one trainer, and the jth trainer can be matched with at most one player.
Return the maximum number of matchings between players and trainers that satisfy these conditions.
Example 1:
Input: players = [4,7,9], trainers = [8,2,5,8] Output: 2 Explanation: One of the ways we can form two matchings is as follows: - players[0] can be matched with trainers[0] since 4 <= 8. - players[1] can be matched with trainers[3] since 7 <= 8. It can be proven that 2 is the maximum number of matchings that can be formed.
Example 2:
Input: players = [1,1,1], trainers = [10] Output: 1 Explanation: The trainer can be matched with any of the 3 players. Each player can only be matched with one trainer, so the maximum answer is 1.
Constraints:
1 <= players.length, trainers.length <= 1051 <= players[i], trainers[j] <= 109
Note: This question is the same as 445: Assign Cookies.
Approach Overview
Problem Overview: You have two arrays: players representing player ability and trainers representing trainer capacity. A trainer can train one player only if trainer >= player. Each trainer can be used once. The task is to maximize the number of valid player‑trainer matches.
Approach 1: Two-Pointer Technique after Sorting (O(n log n) time, O(1) extra space)
Sort both arrays, then greedily match the smallest available player with the smallest trainer that can handle them. Use two pointers: one for players and one for trainers. If the current trainer capacity is greater than or equal to the player's ability, record a match and move both pointers. Otherwise move only the trainer pointer to search for a stronger trainer. Sorting ensures you never waste a high-capacity trainer on a weak player when a smaller trainer could work. The algorithm performs a single linear scan after sorting, making it efficient and easy to implement using concepts from sorting, two pointers, and greedy algorithms.
Approach 2: Greedy Matching with Ordered Sets (O(n log n) time, O(n) space)
Insert all trainer capacities into an ordered structure such as a multiset (C++) or bisect/sorted container style structure in Python. Iterate through players and find the smallest trainer whose capacity is at least the player's ability using a lower_bound-style lookup. If such a trainer exists, match them and remove that trainer from the set so it cannot be reused. Each lookup and removal takes O(log n), and you perform it for each player. This approach is useful when you want explicit control over matching or when processing players in a custom order.
The key greedy insight is that weaker players should consume the weakest valid trainers. Assigning a large-capacity trainer too early can block a future player who actually needs that capacity. Sorting or ordered lookups guarantee that each match uses the smallest feasible trainer.
Recommended for interviews: The sorted two-pointer approach is the expected solution. It is short, easy to reason about, and clearly demonstrates greedy thinking combined with efficient iteration over arrays. Interviewers typically expect O(n log n) due to sorting, followed by a linear scan. The ordered set method shows deeper knowledge of data structures but is rarely necessary unless the problem introduces dynamic updates.
Approach 1: Two-Pointer Technique after Sorting
This approach involves sorting both the players and trainers arrays. Then, using a two-pointer method, we try to find the maximum number of pairings possible. We move pointers on players and trainers based on the comparison between the player's ability and the trainer's capacity.
The solution starts by sorting both arrays, players and trainers. Then, it utilizes two pointers, one for each array. If the current player's ability is less than or equal to the current trainer's capacity, a match is counted, and both pointers advance. If not, only the trainer's pointer advances to find a suitable trainer.
Complexity
Time Complexity: O(n log n + m log m), where n and m are the lengths of the players and trainers arrays respectively, due to sorting.
Space Complexity: O(1), as no additional space proportional to input size is used.
Approach 2: Greedy Matching with Sets
This approach is based on using a greedy algorithm with the help of sets or heaps to facilitate a similar matching process. By leveraging the properties of sets, we can efficiently manage player-trainer pairings as we iterate through sorted lists.
In this C++ solution, we use a multiset to store trainers. We sort players and use the lower_bound function to find the minimum capacity trainer for each player efficiently. Once a match is found, the trainer is removed from the set.
Complexity
Time Complexity: O(n log n + m log m) for sorting and finding matches.
Space Complexity: O(m) for storing trainers.
Approach 3: Greedy + Two Pointers
According to the problem description, each athlete should be matched with the trainer whose ability value is as close as possible. Therefore, we can sort the ability values of both athletes and trainers, and then use the two-pointer method for matching.
We use two pointers i and j to point to the arrays of athletes and trainers, respectively, both initially pointing to the start of the arrays. Then we traverse the ability values of the athletes one by one. If the current trainer's ability value is less than the current athlete's ability value, we move the trainer's pointer to the right by one position until we find a trainer whose ability value is greater than or equal to the current athlete's. If no such trainer is found, it means the current athlete cannot be matched with any trainer, and we return the current athlete's index. Otherwise, we can match the current athlete with the trainer, and then move both pointers to the right by one position. Continue this process until all athletes have been traversed.
If we traverse all athletes, it means all athletes can be matched with trainers, and we return the number of athletes.
The time complexity is O(m times log m + n times log n), and the space complexity is O(max(log m, log n)). Here, m and n are the numbers of athletes and trainers, respectively.
Complexity Comparison
| Approach | Complexity |
|---|---|
| Two-Pointer Technique after Sorting | Time Complexity: O(n log n + m log m), where n and m are the lengths of the |
| Greedy Matching with Sets | Time Complexity: O(n log n + m log m) for sorting and finding matches. |
| Greedy + Two Pointers | — |
Detailed Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Two-Pointer Technique after Sorting | O(n log n) | O(1) | Best general solution; simple greedy matching after sorting both arrays |
| Greedy Matching with Ordered Set | O(n log n) | O(n) | Useful when trainers must be searched dynamically using lower_bound lookups |
Video Solution
Maximum Matching of Players With Trainers | Simple Intuition | Leetcode 2410 | codestorywithMIK • codestorywithMIK • 4,154 views views
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