<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:
        '''
        No completely out of the blue algorithms?
        brute force -> O(n^2)
        '''
        n = len(arr)
        left = [float('inf')]*n

        seen = {0:-1}
        curr, res = 0, float('inf')
        for i in range(n):
            curr += arr[i]
            if curr-target in seen:
                res = min(res,i-seen[curr-target])
            left[i] = res
            seen[curr] = i

        right = [float('inf')]*n

        seen = {0:n}
        curr, res = 0, float('inf')
        for i in range(n-1,-1,-1):
            curr += arr[i]
            if curr-target in seen:
                res = min(res,seen[curr-target]-i)
            right[i] = res
            seen[curr] = i

        res = float('inf')
        for i in range(n-1):
            res = min(res,left[i]+right[i+1])

        if res == float('inf'):
            return -1

        return res

        
        

        



        
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