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Diffstat (limited to 'max_match_players_trainers.c')
-rw-r--r-- | max_match_players_trainers.c | 35 |
1 files changed, 35 insertions, 0 deletions
diff --git a/max_match_players_trainers.c b/max_match_players_trainers.c new file mode 100644 index 0000000..e612986 --- /dev/null +++ b/max_match_players_trainers.c @@ -0,0 +1,35 @@ +// 2410. Maximum Matching of Players With Trainers +#include "leetcode.h" + +/* + * you are given a 0-indexed integer array 'players' where 'players[i]' + * represents the ability of the i'th player. you are also given a 0-indexed + * integer array 'trainers', where 'trainers[j]' represents the j'th capacity + * of the trainer. the i'th player can match with the j'th trainer if the + * players ability is less than or equal to the trainers training capacity. + * additionally, the i'th player can matchedwith at most one trainer, and the + * j'th trainer can be matched with at most one player. return the maximum + * number of matchings between 'players' and 'trainers' that satisfy these + * conditions. + */ + +int cmp(const void *a, const void *b) { return (*(int *)a) - (*(int *)b); } + +int matchPlayersAndTrainers(int *players, int playersSize, int *trainers, + int trainersSize) { + qsort(players, playersSize, sizeof(int), cmp); + qsort(trainers, trainersSize, sizeof(int), cmp); + int cnt = 0; + for (int i = 0; i < trainersSize && cnt < playersSize; ++i) + cnt += players[cnt] <= trainers[i]; + return cnt; +} + +int main() { + int p1[] = {4, 7, 9}, t1[] = {8, 2, 5, 8}; + int p2[] = {1, 1, 1}, t2[] = {10}; + printf("%d\n", matchPlayersAndTrainers(p1, ARRAY_SIZE(p1), t1, + ARRAY_SIZE(t1))); // expect: 2 + printf("%d\n", matchPlayersAndTrainers(p2, ARRAY_SIZE(p2), t2, + ARRAY_SIZE(t2))); // expect: 1 +} |