<p>You are given an array of integers <code>arr</code> and an integer <code>target</code>.</p>

<p>You have to find <strong>two non-overlapping sub-arrays</strong> of <code>arr</code> each with a sum equal <code>target</code>. There can be multiple answers so you have to find an answer where the sum of the lengths of the two sub-arrays is <strong>minimum</strong>.</p>

<p>Return <em>the minimum sum of the lengths</em> of the two required sub-arrays, or return <code>-1</code> if you cannot find such two sub-arrays.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [3,2,2,4,3], target = 3
<strong>Output:</strong> 2
<strong>Explanation:</strong> Only two sub-arrays have sum = 3 ([3] and [3]). The sum of their lengths is 2.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [7,3,4,7], target = 7
<strong>Output:</strong> 2
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 3:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= arr[i] &lt;= 1000</code></li>
	<li><code>1 &lt;= target &lt;= 10<sup>8</sup></code></li>
</ul>

from typing import List
import collections

class Solution:
    def minSumOfLengths(self, arr: List[int], target: int) -> int:
        # Helper function to calculate the minimum subarray lengths with sum equal to target
        def get_sub_arrays(arr):
            lookup = collections.defaultdict(int)  # Dictionary to store the last index of each running sum
            running_sum = 0  # Initialize running sum
            dp = [float('inf')] * len(arr)  # Initialize dp array with infinity

            for i, num in enumerate(arr):
                running_sum += num  # Update running sum
                if running_sum == target:
                    dp[i] = i + 1  # If running sum equals target, update dp with current subarray length
                elif running_sum - target in lookup:
                    dp[i] = i - lookup[running_sum - target] + 1  # Update dp with subarray length if found
                lookup[running_sum] = i + 1  # Update the last index of the current running sum
                if i > 0:
                    dp[i] = min(dp[i], dp[i - 1])  # Ensure dp[i] is the minimum length found so far
            return dp

        # Calculate minimum subarray lengths from the left
        dp_left = get_sub_arrays(arr)
        # Calculate minimum subarray lengths from the right
        dp_right = get_sub_arrays(arr[::-1])[::-1]

        ans = float('inf')  # Initialize answer with infinity
        for i in range(1, len(arr)):
            # Calculate the minimum sum of lengths of two non-overlapping subarrays
            ans = min(ans, dp_left[i - 1] + dp_right[i])

        # Return the minimum sum of lengths if found, otherwise return -1
        return ans if ans != float('inf') else -1
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