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""" | ||
Given an array of integers, find two non-overlapping subarrays which have the largest sum. | ||
The number in each subarray should be contiguous. | ||
Return the largest sum. | ||
Note | ||
The subarray should contain at least one number | ||
Example | ||
For given [1, 3, -1, 2, -1, 2], the two subarrays are [1, 3] and [2, -1, 2] or [1, 3, -1, 2] and [2], they both have | ||
the largest sum 7. | ||
Challenge | ||
Can you do it in time complexity O(n) ? | ||
""" | ||
__author__ = 'Danyang' | ||
class Solution: | ||
def maxTwoSubArrays(self, nums): | ||
""" | ||
dp max subarray | ||
fi for subarry that ends WITH OR BEFORE i | ||
f1 for forward sweeping | ||
f2 for backward sweeping | ||
:param nums: A list of integers | ||
:return: An integer denotes the sum of max two non-overlapping subarrays | ||
""" | ||
n = len(nums) | ||
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f = [[-1<<31 for _ in xrange(n+1)] for _ in xrange(2)] | ||
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cur = 0 | ||
for i in xrange(1, n+1): | ||
cur += nums[i-1] | ||
f[0][i] = max(nums[i-1], f[0][i-1], cur) | ||
if cur<0: | ||
cur = 0 | ||
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cur = 0 | ||
for i in xrange(n-1, -1, -1): | ||
cur += nums[i] | ||
f[1][i] = max(nums[i], f[1][i+1], cur) | ||
if cur<0: | ||
cur = 0 | ||
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maxa = -1<<31 | ||
for i in xrange(1, n): | ||
maxa = max(maxa, f[0][i]+f[1][i]) | ||
return maxa | ||
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if __name__=="__main__": | ||
print Solution().maxTwoSubArrays([1, 3, -1, 2, -1, 2]) | ||
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