LC-2468 Hard LeetCode

2468. Split Message Based on Limit

Read the full problem statement on LeetCode.

Difficulty: hard Acceptance: 43% Topics: String, Binary Search, Enumeration

Reading material

Strings 12 min Binary Search 12 min Backtracking 14 min

Reference solution (spoiler · python)

# Time:  O(n + rlogr), r is the number of messages
# Space: O(1)

# brute force, linear search (binary search doesn't work)
class Solution(object):
    def splitMessage(self, message, limit):
        """
        :type message: str
        :type limit: int
        :rtype: List[str]
        """
        cnt, l, total, base = 1, 1, len(message)+1, 1
        while 3+l*2 < limit:
            if total+(3+l)*cnt <= limit*cnt:
                break
            cnt += 1
            if cnt == base*10:
                l += 1
                base *= 10
            total += l
        if 3+l*2 >= limit:
            return []
        result = []
        j = 0
        for i in xrange(cnt):
            l = limit-(3+len(str(i+1))+len(str(cnt)))
            result.append("%s<%s/%s>"%(message[j:j+l], i+1, cnt))
            j += l
        return result

Solution from kamyu104/LeetCode-Solutions · MIT