Simple NLogN to warm up your day!
Sorting both arrays (NLogN), then a linear scan looking for the matching pairs. Doesn't get more NLogN than this. Code is down below friends, cheers, ACC.
Maximum Matching of Players With Trainers - LeetCode
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 <= 105
1 <= players[i], trainers[j] <= 109
public int MatchPlayersAndTrainers(int[] players, int[] trainers) { Array.Sort(players); Array.Sort(trainers); int pIndex = 0; int tIndex = 0; int retVal = 0; while (pIndex < players.Length && tIndex < trainers.Length) { if (players[pIndex] <= trainers[tIndex]) { retVal++; pIndex++; tIndex++; } else { tIndex++; } } return retVal; }
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