689. Maximum Sum of 3 Non-Overlapping Subarrays

In a given array nums of positive integers, find three non-overlapping subarrays with maximum sum.

Each subarray will be of size k, and we want to maximize the sum of all 3*k entries.

Return the result as a list of indices representing the starting position of each interval (0-indexed). If there are multiple answers, return the lexicographically smallest one.

Example:

Input: [1,2,1,2,6,7,5,1], 2
Output: [0, 3, 5]
Explanation: Subarrays [1, 2], [2, 6], [7, 5] correspond to the starting indices [0, 3, 5].
We could have also taken [2, 1], but an answer of [1, 3, 5] would be lexicographically larger.

Note:

• nums.length will be between 1 and 20000.
• nums[i] will be between 1 and 65535.
• k will be between 1 and floor(nums.length / 3).

这题是FB最近的高频题，也是很经典的题目， 看到subarray的题目无非是dp或者two pointer的题目，但是这题看数组长度非常长，按普通区间DP或者3个pointer这种，复杂度为O(n^2),显然不合适。这题的解法充分利用了3个子数组这个条件，所以可以分开成左中右3个来考虑。用left[i]求0-i区间长度为k的最大和子数组的开始index，right[i] 求[i:len(nums)]区间内长度为k的最大和数组的开始index。之后我们针对中间数组，可以同步考虑左右边的情况，代码如下：

class Solution(object):
    def maxSumOfThreeSubarrays(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        # simplified version, left and right arrays have the same length with nums
        n = len(nums)
        if n < 3*k:
            return []
        presum = [0]
        presum[j] - presum[i] = sum(nums[i:j])
        for i in nums:
            presum.append(presum[-1]+i)
        # left[i], from 0-i, largest sum subarray of length k's start index
        left = [0] * n
        # right[i], from i-n, largest sum subarray of length k's start index
        right = [n-k] * n
        
        for i in xrange(k-1, n-2*k):
            if i == k-1:
                maxsum = presum[i+1] - presum[0]
                left[i] = 0
            else:
                if presum[i+1] - presum[i-k+1] > maxsum:
                    maxsum = presum[i+1] - presum[i-k+1]
                    left[i] = i - k + 1 
                else:
                    left[i] = left[i-1]
                    
        for i in xrange(n-k, 2*k-1, -1):
            if i == n-k:
                maxsum = presum[n] - presum[n-k]
                right[i] = n - k
            else:
                if presum[i+k] - presum[i] > maxsum:
                    maxsum = presum[i+k] - presum[i]
                    right[i] = i
                else:
                    right[i] = right[i+1]
        res = []
        maxsum = 0
        for i in xrange(k, n-2*k+1):
            l = left[i-1]
            r= right[i+k]
            if presum[i+k] - presum[i] + presum[l+k] - presum[l] + presum[r+k] - presum[r] > maxsum:
                maxsum = presum[i+k] - presum[i] + presum[l+k] - presum[l] + presum[r+k] - presum[r]
                res = [l, i, r]
        return res        

注意这里为了方便left和right都取成了和nums一样的长度，但是只有其中一部分有意义。