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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);
+}