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Problem

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

A subarray is a contiguous part of an array.

 

Example 1:

Input: nums = [-2,1,-3,4,-1,2,1,-5,4]Output: 6Explanation: [4,-1,2,1] has the largest sum = 6.

Example 2:

Input: nums = [1]Output: 1

Example 3:

Input: nums = [5,4,-1,7,8]Output: 23

 

Constraints:

  • 1 <= nums.length <= 3 * 104
  • -105 <= nums[i] <= 105

 

Follow up: If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

Sol

def helper(l):
    if len(l) == 0:
        return 0
    ret = tmp = l[0]
    for n in l[1:]:
        tmp += n
        ret = max(ret, tmp)
    return ret
class Solution:
    def maxSubArray(self, l: List[int]) -> int:
        if not l:
            return 0
        if len(l) == 1:
            return l[0]
        else:
            if len(l) % 2 == 0:
                mid = len(l) // 2
                onlyLeft = self.maxSubArray(l[:mid])
                onlyRight = self.maxSubArray(l[mid:])
                center = 0
                left = helper(l[:mid][::-1])
                right = helper(l[mid:])
            else:
                mid = len(l) // 2
                onlyLeft = self.maxSubArray(l[:mid+1])
                onlyRight = self.maxSubArray(l[mid:])
                center = l[mid]
                left = helper(l[:mid][::-1])
                right = helper(l[mid+1:])
        return max([center + left + right, onlyLeft, onlyRight])

Sol2

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        ret = tmp = nums[0]
        for n in nums[1:]:
            tmp = max(n, n+tmp)
            ret = max(ret, tmp)
        return ret