weekly contest

Author: A-stick-bug

3438. Find Valid Pair of Adjacent Digits in String.py

@@ -0,0 +1,10 @@
+def findValidPair(s: str) -> str:
+    for i in range(1, len(s)):
+        if s[i] != s[i - 1] and s.count(s[i]) == int(s[i]) and s.count(s[i - 1]) == int(s[i - 1]):
+            return s[i - 1] + s[i]
+    return ""
+
+
+print(findValidPair("2523533"))
+print(findValidPair("221"))
+print(findValidPair("22"))

3439. Reschedule Meetings for Maximum Free Time I.py

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+# https://leetcode.com/problems/reschedule-meetings-for-maximum-free-time-i
+# Sliding window approach, window size = k + 1
+# The max contiguous we can get from a window is (length - sum of interval lengths inside)
+# We can think of this as putting all intervals in the window next to each other
+
+
+def maxFreeTime(eventTime: int, k: int, startTime: list[int], endTime: list[int]) -> int:
+    removed = 0
+    startTime = [0] + startTime + [eventTime]
+    endTime = [0] + endTime + [eventTime]
+
+    best = 0
+    for i, (s, e) in enumerate(zip(startTime, endTime)):
+        removed += e - s
+
+        li = i - (k + 1)
+        if li >= 0:
+            removed -= endTime[li] - startTime[li]
+            cur_le = e - endTime[li] - removed
+            best = max(best, cur_le)
+
+    return best
+
+
+print(maxFreeTime(eventTime=19, k=2, startTime=[1, 5, 11, 14], endTime=[3, 7, 13, 17]))
+print(maxFreeTime(eventTime=5, k=1, startTime=[1, 3], endTime=[2, 5]))
+print(maxFreeTime(eventTime=10, k=1, startTime=[0, 2, 9], endTime=[1, 4, 10]))
+print(maxFreeTime(eventTime=5, k=2, startTime=[0, 1, 2, 3, 4], endTime=[1, 2, 3, 4, 5]))

3440. Reschedule Meetings for Maximum Free Time II.py

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+# https://leetcode.com/problems/reschedule-meetings-for-maximum-free-time-ii
+# For each interval, check if we can move it somewhere else
+# We can od this by maintaining all 'gaps' in a SortedList
+
+from sortedcontainers import SortedList
+
+
+def maxFreeTime(eventTime: int, startTime: list[int], endTime: list[int]) -> int:
+    startTime = [0, 0] + startTime + [eventTime, eventTime]  # padding
+    endTime = [0, 0] + endTime + [eventTime, eventTime]
+
+    gaps = SortedList([startTime[i] - endTime[i - 1] for i in range(1, len(endTime))])
+
+    best = 0
+    for i in range(1, len(endTime) - 1):
+        l = startTime[i] - endTime[i - 1]  # remove gaps right next to this interval
+        r = startTime[i + 1] - endTime[i]
+        gaps.remove(l)
+        gaps.remove(r)
+
+        le = endTime[i] - startTime[i]
+        if gaps and le <= gaps[-1]:  # we can move this interval somewhere else
+            cur_le = startTime[i + 1] - endTime[i - 1]
+        else:  # we cannot move it somewhere else, instead push it to the left to make as much space as possible
+            cur_le = startTime[i + 1] - endTime[i - 1] - (endTime[i] - startTime[i])
+        best = max(best, cur_le)
+
+        gaps.add(l)  # add the gaps back now that we are done
+        gaps.add(r)
+    return best
+
+
+print(maxFreeTime(eventTime=19, startTime=[1, 5, 11, 14], endTime=[3, 8, 13, 17]))
+
+print(maxFreeTime(eventTime=5, startTime=[1, 3], endTime=[2, 5]))
+print(maxFreeTime(eventTime=10, startTime=[0, 7, 9], endTime=[1, 8, 10]))
+print(maxFreeTime(eventTime=10, startTime=[0, 3, 7, 9], endTime=[1, 4, 8, 10]))
+print(maxFreeTime(eventTime=5, startTime=[0, 1, 2, 3, 4], endTime=[1, 2, 3, 4, 5]))