<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:
        def get_sub_arrays( arr ):
            lookup = collections.defaultdict(int)
            running_sum = 0
            dp = [float('inf')] * len(arr)
            
            for i, num in enumerate(arr):
                running_sum += num
                if running_sum == target:
                    dp[i] = i - 0 + 1
                elif running_sum - target in lookup:
                    dp[i] = i - lookup[running_sum - target] + 1
                lookup[running_sum] = i+1
                dp[i] = min( dp[i-1], dp[i] )
            return dp
        
        dp_left = get_sub_arrays( arr )                     # from front
        dp_right = get_sub_arrays( arr[::-1] )[::-1]        # from backwards
        
        ans = float('inf')
        for i in range( 1, len(arr) ):
            ans = min( ans, dp_left[i-1] + dp_right[i] )
        return ans if( ans != float('inf') ) else -1
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