<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>

class Solution:
    def minSumOfLengths(self, arr: List[int], target: int) -> int:
        n = len(arr)
        prev_pos_lookup = [-1] * n
        DP = [inf] * n
        pos_lookup = {0: -1}
        for pos, cumulative in enumerate(accumulate(arr)):
            prev = cumulative - target
            if prev in pos_lookup:
                prev_pos = pos_lookup[prev]
                prev_pos_lookup[pos] = prev_pos
                DP[pos] = pos - prev_pos
            if pos:
                DP[pos] = min(DP[pos], DP[pos - 1])
            pos_lookup[cumulative] = pos
        
        res = inf
        for pos in range(n - 1, -1, -1):
            prev_pos = prev_pos_lookup[pos]
            if prev_pos == -1:
                continue
            res = min(res, pos - prev_pos + DP[prev_pos])
        return res if res != inf else -1
40900