human
I found an interesting problem on https://leetcode.com/problems/find-two-non-overlapping-sub-arrays-each-with-target-sum/:
Given an array of integers arr and an integer target.
You have to find two non-overlapping sub-arrays of arr each with sum equal target. There can be multiple answers so you have to find an answer where the sum of the lengths of the two sub-arrays is minimum.
Return the minimum sum of the lengths of the two required sub-arrays, or return -1 if you cannot find such two sub-arrays.
 
Example 1:
Input: arr = [3,2,2,4,3], target = 3
Output: 2
Explanation: Only two sub-arrays have sum = 3 ([3] and [3]). The sum of their lengths is 2.

Example 2:
Input: arr = [7,3,4,7], target = 7
Output: 2
Explanation: Although we have three non-overlapping sub-arrays of sum = 7 ([7], [3,4] and [7]), but we will choose the first and third sub-arrays as the sum of their lengths is 2.

Example 3:
Input: arr = [4,3,2,6,2,3,4], target = 6
Output: -1
Explanation: We have only one sub-array of sum = 6.

Example 4:
Input: arr = [5,5,4,4,5], target = 3
Output: -1
Explanation: We cannot find a sub-array of sum = 3.

Example 5:
Input: arr = [3,1,1,1,5,1,2,1], target = 3
Output: 3
Explanation: Note that sub-arrays [1,2] and [2,1] cannot be an answer because they overlap.

 
Constraints:

1 <= arr.length <= 10^5
1 <= arr[i] <= 1000
1 <= target <= 10^8

I tried it in Python, but could not do it. Can you solve it?

 ... Write me a Python code to find two non-overlapping sub-arrays of arr each with sum equal target?
assistant
#Here is Python code to find two non-overlapping sub-arrays of arr each with sum equal target.

class Solution:
  # 872 ms
  def minSumOfLengths(self, arr, target):
    result = inf = 2**31-1
    i = window = 0
    premin = [inf] * len(arr)

    # i: window start, j: window end
    for j, num in enumerate(arr):
      window += num
      while window > target:
        window -= arr[i]
        i += 1
      if window == target:
        # curr: length
        curr = j - i + 1
        if result > curr + premin[i - 1]:
            result = curr + premin[i - 1]
        premin[j] = curr if curr < premin[j - 1] else premin[j - 1]
        
      else:
        premin[j] = premin[j - 1]

    return result if result < inf else -1