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Diffstat (limited to 'earliest_latest_rounds_players_compete.c')
-rw-r--r-- | earliest_latest_rounds_players_compete.c | 61 |
1 files changed, 61 insertions, 0 deletions
diff --git a/earliest_latest_rounds_players_compete.c b/earliest_latest_rounds_players_compete.c new file mode 100644 index 0000000..3784964 --- /dev/null +++ b/earliest_latest_rounds_players_compete.c @@ -0,0 +1,61 @@ +// 1900. The Earliest and Latest Rounds Where Players Compete +#include "leetcode.h" + +/* + * there is a tournament where 'n' players are participating. the players are + * standing a single row and are numbered from 1 to 'n' based on their initial + * standing position. the tournament consists of multiple rounds. in each round, + * the i'th player from the front of the row competes against the i'th player + * from the end of the row, and the winner advances to the next round. when the + * number of players is odd for the current round, te player is in the middle + * automatically advances to the next round. after each round is over, the + * winners are lined back up in the row based on the original ordering assigned + * to them initially. the players numbered 'firstPlayer' and 'secondPlayer' are + * the best in the tournament. they can win agsainst any other player beofre + * they compete against each other. if any two other players compete against + * each other, either of them might win, and thus you may choose the outcome of + * this round. + */ + +int *earliestAndLatest(int n, int firstPlayer, int secondPlayer, + int *returnSize) { + *returnSize = 2; + int *ans = (int *)malloc(*returnSize * sizeof(int)); + int first = firstPlayer, second = secondPlayer; + if (first + second == n + 1) { + ans[0] = ans[1] = 1; + return ans; + } + if (first + second > n + 1) { + int tmp = first; + first = n + 1 - second; + second = n + 1 - tmp; + } + int pre = 1; + int pos = (32 - __builtin_clz(n - 1) < n + 1 - second) + ? 32 - __builtin_clz(n - 1) + : n + 1 - second; + if (first + second == n) { + if (!(first % 2)) + pre = (first + 2 < second) ? 2 : 1 + __builtin_ctz(first); + } else if (first + 1 == second) { + pre = 32 - __builtin_clz((n - 1) / (first + second - 1)); + pre += __builtin_ctz((n - 1) >> pre); + } else if (first + second <= (n + 1) / 2) + pre = 32 - __builtin_clz((n - 1) / (first + second - 1)); + ans[0] = pre + 1; + ans[1] = pos; + return ans; +} + +int main() { + int rs1, *eal1 = earliestAndLatest(11, 2, 4, &rs1); + int rs2, *eal2 = earliestAndLatest(5, 1, 5, &rs2); + for (int i = 0; i < rs1; i++) + printf("%d ", eal1[i]); // expect: 3 4 + printf("\n"); + for (int i = 0; i < rs2; i++) + printf("%d ", eal2[i]); // expect: 1 1 + printf("\n"); + free(eal1), free(eal2); +} |