1977. Number of Ways to Separate Numbers

# Time:  O(n^2)
# Space: O(n^2)

class Solution(object):
    def numberOfCombinations(self, num):
        """
        :type num: str
        :rtype: int
        """
        MOD = 10**9+7
        def find_longest_common_prefix(num):
            lcp = [[0]*(len(num)+1) for _ in xrange(len(num)+1)]  # lcp[i][j]: longest length of the common prefix which starts at num[i], num[j]
            for i in reversed(xrange(len(lcp)-1)):
                for j in reversed(xrange(len(lcp[0])-1)):
                    if num[i] == num[j]:
                        lcp[i][j] = lcp[i+1][j+1]+1
            return lcp

        def is_less_or_equal_to_with_same_length(num, lcp, i, j, l):
            return lcp[i][j] >= l or num[i+lcp[i][j]] < num[j+lcp[i][j]]

        lcp = find_longest_common_prefix(num)
        dp = [[0]*len(num) for _ in xrange(len(num))]  # dp[i][l]: the count of numbers ending at num[i], where the length of the last number is l+1
        dp[0][0] = int(num[0] != '0')
        for i in xrange(1, len(num)):
            dp[i][i] = dp[i-1][i-1]
            if num[i] == '0':
                continue
            accu = 0
            for l in xrange(len(num)-i+1):
                ni = i+l-1
                dp[ni][l-1] = accu  # accumulated count where the length of the second to last number ending at num[i-1] is shorter than the length of the last number ending at num[i+l-1]
                if i-l < 0:
                    continue
                if num[i-l] != '0' and is_less_or_equal_to_with_same_length(num, lcp, i-l, i, l):
                    dp[ni][l-1] = (dp[ni][l-1] + dp[i-1][l-1]) % MOD
                accu = (accu + dp[i-1][l-1]) % MOD
        return reduce(lambda total, x: (total+x)%MOD, dp[-1], 0)

Solution from kamyu104/LeetCode-Solutions · MIT