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// 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);
}
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