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create-maximum-number.py
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create-maximum-number.py
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# Time: O(k * (m + n + k)) ~ O(k * (m + n + k^2))
# Space: O(m + n + k^2)
#
# Given two arrays of length m and n with digits 0-9 representing two numbers.
# Create the maximum number of length k <= m + n from digits of the two.
# The relative order of the digits from the same array must be preserved.
# Return an array of the k digits. You should try to optimize your time
# and space complexity.
#
# Example 1:
# nums1 = [3, 4, 6, 5]
# nums2 = [9, 1, 2, 5, 8, 3]
# k = 5
# return [9, 8, 6, 5, 3]
#
# Example 2:
# nums1 = [6, 7]
# nums2 = [6, 0, 4]
# k = 5
# return [6, 7, 6, 0, 4]
#
# Example 3:
# nums1 = [3, 9]
# nums2 = [8, 9]
# k = 3
# return [9, 8, 9]
#
# DP + Greedy solution. (280ms)
class Solution(object):
def maxNumber(self, nums1, nums2, k):
"""
:type nums1: List[int]
:type nums2: List[int]
:type k: int
:rtype: List[int]
"""
def get_max_digits(nums, start, end, max_digits):
max_digits[end] = max_digit(nums, end)
for i in reversed(xrange(start, end)):
max_digits[i] = delete_digit(max_digits[i + 1])
def max_digit(nums, k):
drop = len(nums) - k
res = []
for num in nums:
while drop and res and res[-1] < num:
res.pop()
drop -= 1
res.append(num)
return res[:k]
def delete_digit(nums):
res = list(nums)
for i in xrange(len(res)):
if i == len(res) - 1 or res[i] < res[i + 1]:
res = res[:i] + res[i+1:]
break
return res
def merge(a, b):
return [max(a, b).pop(0) for _ in xrange(len(a)+len(b))]
m, n = len(nums1), len(nums2)
max_digits1, max_digits2 = [[] for _ in xrange(k + 1)], [[] for _ in xrange(k + 1)]
get_max_digits(nums1, max(0, k - n), min(k, m), max_digits1)
get_max_digits(nums2, max(0, k - m), min(k, n), max_digits2)
return max(merge(max_digits1[i], max_digits2[k-i]) \
for i in xrange(max(0, k - n), min(k, m) + 1))