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LeetCode 2104 Sum of Subarray Ranges

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LeetCode



You are given an integer array nums. The range of a subarray of nums is the difference between the largest and smallest element in the subarray.

Return the sum of all subarray ranges of nums.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [1,2,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0
[2], range = 2 - 2 = 0
[3], range = 3 - 3 = 0
[1,2], range = 2 - 1 = 1
[2,3], range = 3 - 2 = 1
[1,2,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 1 + 1 + 2 = 4.


Example 2:

Input: nums = [1,3,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0
[3], range = 3 - 3 = 0
[3], range = 3 - 3 = 0
[1,3], range = 3 - 1 = 2
[3,3], range = 3 - 3 = 0
[1,3,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 2 + 0 + 2 = 4.


Example 3:

Input: nums = [4,-2,-3,4,1]
Output: 59
Explanation: The sum of all subarray ranges of nums is 59.


Constraints:

1 <= nums.length <= 1000 -10<sup>9</sup> <= nums[i] <= 10<sup>9</sup>

题目大意：

求所有子数组的最大值最小值之差的和

Stack算法思路：

参考Leetcode 907，分别求子数组最小值的相反数，子数组的最大值，这两个值的和即为所求

注意事项：

  1. 最小值用递增栈，最大值用递减栈

Python代码：

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def subArrayRanges(self, nums: List[int]) -> int:
	arr = list(nums)
	arr.insert(0, -sys.maxsize)
	arr.append(-sys.maxsize)
	stack, res, = [], 0
	for i in range(len(arr)):
		while stack and arr[i] < arr[stack[-1]]:
			prev_idx = stack.pop()
			res -= arr[prev_idx] * (prev_idx - stack[-1]) * (i - prev_idx)
		stack.append(i)

	arr = list(nums)
	arr.insert(0, sys.maxsize)
	arr.append(sys.maxsize)

	for i in range(len(arr)):
		while stack and arr[i] > arr[stack[-1]]:
			prev_idx = stack.pop()
			res += arr[prev_idx] * (prev_idx - stack[-1]) * (i - prev_idx)
		stack.append(i)
	return res

算法分析：

时间复杂度为O(n)，空间复杂度O(n)

累计和算法II解题思路：

比暴力法稍优，两重循环覆盖所有子数组[i, j]，每轮循环得到最大最小值，然后O(1)内求该区间内所有最大最小值差值和。

Python代码：

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def subArrayRanges(self, nums: List[int]) -> int:
	sum = 0
	for i in range(len(nums) - 1):
		min_value, max_value = nums[i], nums[i]
		for j in range(i + 1, len(nums)):
			min_value = min(min_value, nums[j])
			max_value = max(max_value, nums[j])
			sum += max_value - min_value
	return sum

算法分析：

时间复杂度为O(n^2)，空间复杂度O(1)

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