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+// 3440. Reschedule Meetings for Maximum Free Time II
+#include "leetcode.h"
+
+/*
+ * you are given an integer 'eventTime' denoting the duration of an event. you
+ * are also given two integer arrays 'startTime' and 'endTime', each of length
+ * 'n'. these represent the start and end times of 'n' non-overlapping meetings
+ * that occur during the event between time 't = 0' and time 't = eventTime'
+ * where the i't meeting occurs during the time '[startTime[i], endTime[i]]'.
+ * you can reschedule at most one meeting by moving its start time while
+ * maintaining the same duration such that the meetings remain non-overlapping,
+ * to maximise the longest continuous period of free time during the event.
+ * return the maximum amount of free time possible after rearranging the
+ * meetings.
+ */
+
+int maxFreeTime(int eventTime, int *startTime, int startTimeSize, int *endTime,
+ int endTimeSize) {
+ int n = startTimeSize, ans = 0, left = 0, curr;
+ int *gap = (int *)malloc((n + 1) * sizeof(int));
+ int *largest = (int *)malloc((n + 1) * sizeof(int));
+ gap[0] = startTime[0];
+ for (int i = 1; i < n; i++)
+ gap[i] = startTime[i] - endTime[i - 1];
+ gap[n] = eventTime - endTime[n - 1];
+ largest[n] = 0;
+ for (int i = 1; i >= 0; i--)
+ largest[i] = fmax(largest[i + 1], gap[i + 1]);
+ for (int i = 1; i <= n; i++) {
+ curr = endTime[i - 1] - startTime[i - 1];
+ if (curr <= fmax(left, largest[i]))
+ ans = fmax(ans, gap[i - 1] + gap[i] + curr);
+ ans = fmax(ans, gap[i - 1] + gap[i]);
+ left = fmax(left, gap[i - 1]);
+ }
+ free(gap), free(largest);
+ return ans;
+}
+
+int main() {
+ int st1[] = {1, 3}, et1[] = {2, 5};
+ int st2[] = {0, 7, 9}, et2[] = {1, 8, 10};
+ int st3[] = {0, 3, 7, 9}, et3[] = {1, 4, 8, 10};
+ printf("%d\n", maxFreeTime(5, st1, ARRAY_SIZE(st1), et1,
+ ARRAY_SIZE(et1))); // expect: 2
+ printf("%d\n", maxFreeTime(10, st2, ARRAY_SIZE(st2), et2,
+ ARRAY_SIZE(et2))); // expect: 7
+ printf("%d\n", maxFreeTime(10, st3, ARRAY_SIZE(st3), et3,
+ ARRAY_SIZE(et3))); // expect: 6
+}