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LC1588 - Sum of All Odd Length Subarrays

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Leetcode: 131/2658 (54+63+14)

Description

Given an array of positive integers arr, return the sum of all possible odd-length subarrays of arr.

A subarray is a contiguous subsequence of the array.

Example 1:

Input: arr = [1,4,2,5,3]
Output: 58
Explanation: The odd-length subarrays of arr and their sums are:
[1] = 1
[4] = 4
[2] = 2
[5] = 5
[3] = 3
[1,4,2] = 7
[4,2,5] = 11
[2,5,3] = 10
[1,4,2,5,3] = 15
If we add all these together we get 1 + 4 + 2 + 5 + 3 + 7 + 11 + 10 + 15 = 58
Example 2:

Input: arr = [1,2]
Output: 3
Explanation: There are only 2 subarrays of odd length, [1] and [2]. Their sum is 3.
Example 3:

Input: arr = [10,11,12]
Output: 66

Constraints:

1 <= arr.length <= 100
1 <= arr[i] <= 1000

Solution

Brute Force

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# O(n^3) time | O(1) space
class Solution:
    def sumOddLengthSubarrays(self, arr: List[int]) -> int:
        result = 0

        for odd in range(1, len(arr)+1, 2):
            cur = 0
            while cur + odd <= len(arr):
                summ = sum(arr[cur:cur+odd])
                cur+=1
                result += summ
        return result

Prefix Sum

  • Prefix sum, num[i] represents the sum of 0 ~ i numbers, then num[i] = num[i] + sum(0 ~ i - 1)
  • Therefore, you only need to find the odd length (denoted as i) that is less than the length of the arr array, and then calculate the difference between the lengths num[j + i] - num[j] in turn.
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# O(n^2) time | O(n) space
class Solution:
    def sumOddLengthSubarrays(self, arr: List[int]) -> int:
        summ = [0]
        res = 0
        for i in arr:
            summ.append(summ[-1]+i)

        for odd in range(1, len(arr)+1, 2):
            left = 0
            while left+odd < len(summ):
                res+= summ[left+odd]-summ[left]
                left+=1
        return res