Problem Description

In this LeetCode problem, you are given two 0-indexed integer arrays: players and trainers. Each entry in the players array represents the ability of a particular player, and each entry in the trainers array represents the training capacity of a particular trainer. A player can be matched with a trainer if the player's ability is less than or equal to the trainer's capacity. It's important to note that each player can only be matched with one trainer, and each trainer can match with only one player. The task is to find the maximum number of such player-trainer pairings where the conditions of matching ability to capacity are met.

Intuition

The intuition behind the solution is straightforward: we want to pair as many players with trainers as we can, prioritizing the pairing of less able players with trainers who have just enough capacity to train them. This ensures that we do not 'waste' a trainer with high capacity on a player with low ability when that trainer could be matched with a more able player instead.

To achieve this, we sort both the players and trainers arrays. Sorting helps us easily compare the least able player with the least capable trainer and move up their respective arrays. If a match is found, both the player and the trainer are effectively removed from the potential pool by moving to the next elements in the arrays (incrementing the index). If no match is found, we move up the trainers array to find the next trainer with a higher capacity that might match the player’s ability.

The result is incrementally built by adding a match whenever we find a capable trainer for a player. We stop when we've either paired all players or run out of trainers. The sum of matches made gives us the maximum number of possible pairings.

Learn more about Greedy, Two Pointers and Sorting patterns.

Solution Approach

The implementation provided uses a simple but effective greedy approach, heavily relying on sorting. The steps involve:

1. Sorting the players array in ascending order, which ensures we start with the player with the lowest ability.

2. Sorting the trainers array in ascending order, which allows us to start with the trainer having the lowest capacity.

The algorithm works with two pointers, one (i) iterating through the players array, and the other (j) through the trainers array.

Here is the implementation broken down:

• Initialize a variable ans to 0 to keep track of the number of matches made.

• Initialize the trainer index pointer j to 0.

• Iterate over each player p in the sorted players list using a for-loop.

  • Within the loop, proceed with an inner while-loop which continues as long as j is less than the length of trainers and the current trainer's capacity trainers[j] is less than the ability of the player p. The purpose of this loop is to find the first trainer in the sorted list whose capacity is sufficient to train the player.

  • After the while-loop, if j is still within the bounds of the trainers array, it means a suitable trainer has been found for the player p. We increment ans to count the match and also increment j to move to the next trainer.

• Once all players have been considered, return the value of ans.

Algorithm and Data Structures:

• Lists: The main data structure used here are lists (players and trainers), which are sorted.

• Greedy approach: The algorithm uses a greedy method by matching each player with the "smallest" available trainer that can train them.

• Two pointers: It uses two pointers to pair players with trainers without revisiting or re-comparing them, thus optimizing the process.

• Sorting: By sorting the arrays, the solution leverages the property that once a player cannot be paired with a current trainer, they cannot be paired with any trainers before that one in the sorted list either.

Patterns used:

• Sorting and Two-pointers: This is a common pattern for efficiently pairing elements from two different sorted lists.

Example Walkthrough

Let's say we have the following players and trainers arrays:

• players: [2, 3, 4]
• trainers: [1, 2, 5]

Following the solution approach:

1. We first sort both the players and trainers arrays:

   After sorting,

   • players: [2, 3, 4] (already sorted)
   • trainers: [1, 2, 5] (already sorted)

2. We initialize ans to 0. This variable will keep track of the successful player-trainer matches.

3. We also initialize the trainer index pointer j to 0.

Now, let's go through each player and try to find a trainer match:

• Player with ability 2 is the first in the players array.

  • We check the trainers in order: trainer 1 cannot train the player because the trainer's capacity is too low.
  • Move to the next trainer (2), and we find a match. Trainer 2 can train the player with ability 2.
  • We increment ans to 1 and move to the next trainer (j becomes 2).

• Player with ability 3 is the next.

  • The trainer in position 2 in the sorted trainers list has a capacity of 5, which is sufficient.
  • We match this player with the trainer.
  • We increment ans to 2, and since there are no more trainers, we move on to the final player.

• Player with ability 4:

  • There are no more trainers to compare since we exceeded the length of the trainers array.

At the end of this process, the value of ans is 2, which signifies that we were able to match two pairs of players and trainers successfully.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code is determined by the sorting operations and the subsequent for-loop with the inner while loop.

1. Sorting both the players and trainers list takes O(nlogn) and O(mlogm), respectively, where n is the number of players and m is the number of trainers.

2. The for-loop iterates over each player, which is O(n). Within this loop, the while loop progresses without resetting, which makes it linear over m elements of trainers in total. In the worst case, all players are checked against all trainers, hence it contributes a maximum of O(m) time complexity.

The combined time complexity of these operations would be O(nlogn) + O(mlogm) + O(n + m). However, since O(nlogn) and O(mlogm) are the dominant terms, the overall time complexity simplifies to O(nlogn + mlogm).

Space Complexity

The space complexity of the code is O(1), which is the additional space required. The sorting happens in place, and no additional data structures are used aside from a few variables for keeping track of indexes and the answer.

Learn more about how to find time and space complexity quickly using problem constraints.