Leetcode 2410: Maximum Matching of Players With Trainers
You are given two arrays: ‘players’ and ’trainers’. The array ‘players’ represents the abilities of different players, and the array ’trainers’ represents the training capacities of various trainers. A player can be matched with a trainer if the player’s ability is less than or equal to the trainer’s capacity. The goal is to find the maximum number of valid matchings between players and trainers such that each player is matched with at most one trainer and vice versa.
Problem
Approach
Steps
Complexity
Input: The input consists of two integer arrays: 'players' and 'trainers'.
Example: players = [5, 6, 8], trainers = [7, 2, 6, 8]
Constraints:
• 1 <= players.length, trainers.length <= 10^5
• 1 <= players[i], trainers[j] <= 10^9
Output: Return an integer that represents the maximum number of valid matchings between players and trainers.
Example: Output: 3
Constraints:
Goal: The goal is to maximize the number of valid pairings by comparing the abilities of players with the training capacities of trainers.
Steps:
• 1. Sort both the players and trainers arrays.
• 2. Use two pointers to try matching the smallest available player with the smallest available trainer that can accommodate them.
• 3. Move through both lists and count the valid pairings, ensuring each player and trainer is matched at most once.
Goal: The solution should efficiently handle the input size up to the upper limits of the constraints.
Steps:
• Sorting both arrays will take O(n log n) time, where n is the size of the arrays.
• Space complexity should be O(1) beyond the input arrays.
Assumptions:
• Players can only be matched with a trainer if their ability is less than or equal to the trainer's capacity.
• Each player and trainer can be matched at most once.
• Input: players = [5, 6, 8], trainers = [7, 2, 6, 8]
• Explanation: After sorting the arrays, players = [5, 6, 8] and trainers = [2, 6, 7, 8]. We can match player 5 with trainer 6, player 6 with trainer 7, and player 8 with trainer 8. Thus, the answer is 3.
• Input: players = [1, 1, 1], trainers = [10]
• Explanation: In this case, all players have the same ability (1), and there is only one trainer with capacity 10. The trainer can be matched with any of the three players, so the maximum number of matches is 1.
Approach: The solution uses sorting and a greedy approach to maximize the number of valid matchings between players and trainers.
Observations:
• The problem can be efficiently solved by sorting both the players and trainers arrays.
• By matching players from the smallest to the largest with the trainers, we can maximize the number of valid pairings.
• After sorting, we can use a two-pointer technique to find the optimal pairings.
Steps:
• 1. Sort the 'players' and 'trainers' arrays in ascending order.
• 2. Use two pointers: one to iterate through the 'players' array and one to iterate through the 'trainers' array.
• 3. If the current player can be matched with the current trainer, increment both pointers and count the match.
• 4. If the player cannot be matched, move to the next trainer.
• 5. Return the total number of matches.
Empty Inputs:
• The input arrays will not be empty as per the problem constraints.
Large Inputs:
• For large inputs, the solution needs to efficiently handle the sorting and matching process.
Special Values:
• If all players have the same ability, the solution should still handle the maximum number of matches correctly.
• If all trainers have much higher capacity than players, then the number of matches will be limited by the number of players.
Constraints:
• The solution should work within the given constraints of the problem.
int matchPlayersAndTrainers(vector<int>& man, vector<int>& master) {
priority_queue<int, vector<int>, greater<int>> pq, qq;
int m = man.size(), n = master.size();
for(int i = 0; i < m; i++) {
pq.push(man[i]);
}
for(int i = 0; i < n; i++) {
qq.push(master[i]);
}
int cnt = 0;
while(!pq.empty() && !qq.empty()) {
while(!qq.empty() && pq.top() > qq.top())
qq.pop();
if(!qq.empty()) {
cnt++;
qq.pop();
pq.pop();
}
}
return cnt;
}
1 : Function Declaration
int matchPlayersAndTrainers(vector<int>& man, vector<int>& master) {
Declares the function `matchPlayersAndTrainers` that takes two vectors representing players and trainers' skills.
2 : Queue Initialization
priority_queue<int, vector<int>, greater<int>> pq, qq;
Creates two priority queues (`pq` for players and `qq` for trainers) to store their skills in ascending order.
3 : Size Calculation
int m = man.size(), n = master.size();
Calculates the size of the `man` and `master` vectors to determine the number of players and trainers.
4 : For Loop - Players
for(int i = 0; i < m; i++) {
Begins a loop to iterate through the players and push their skills into the `pq` priority queue.
5 : Queue Push - Players
pq.push(man[i]);
Pushes the skill of the current player into the `pq` queue.
6 : For Loop - Trainers
for(int i = 0; i < n; i++) {
Begins a loop to iterate through the trainers and push their skills into the `qq` priority queue.
7 : Queue Push - Trainers
qq.push(master[i]);
Pushes the skill of the current trainer into the `qq` queue.
8 : Counter Initialization
int cnt = 0;
Initializes a counter variable `cnt` to keep track of the valid player-trainer pairings.
9 : While Loop - Main Matching
while(!pq.empty() && !qq.empty()) {
Begins a while loop that runs as long as there are players and trainers in their respective queues.
10 : While Loop - Trainer Skill Comparison
while(!qq.empty() && pq.top() > qq.top())
This inner while loop ensures that trainers with lesser skills than the current player are removed from the queue.
11 : Queue Pop - Trainer
qq.pop();
Pops a trainer from the `qq` queue if their skill is less than the player's skill.
12 : If Statement - Valid Match
if(!qq.empty()) {
Checks if there are still trainers left after removing those with insufficient skill to match the current player.
13 : Counter Increment
cnt++;
Increments the counter `cnt` when a valid player-trainer match is found.
14 : Queue Pop - Player
qq.pop();
Pops a trainer from the `qq` queue after a match is made.
15 : Queue Pop - Trainer
pq.pop();
Pops a player from the `pq` queue after a match is made.
16 : Return Statement
return cnt;
Returns the final count of valid player-trainer matches.
Best Case: O(n log n), where n is the size of the larger of the two arrays (players or trainers).
Average Case: O(n log n), as the sorting step dominates the complexity.
Worst Case: O(n log n), since we are always sorting both arrays.
Description: The solution involves sorting both arrays, which takes O(n log n) time.
Best Case: O(1), as no extra space is needed beyond the input arrays.
Worst Case: O(1), as we only use a few additional variables to track the pointers and match counts.
Description: The space complexity is constant, as no additional data structures are used beyond the input arrays.
| LeetCode Solutions Library / DSA Sheets / Course Catalog |
|---|
comments powered by Disqus