<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>


# note: all elements are positive, so we dont need hashmap and can solve with just sliding window + DP (can be greedy)
# use DP to keep track of the length of shortest subarray with sum target until this point
# when we have a subarray with target sum, check if we seen a soln at i-1 then add our current window to it

class Solution:
    def minSumOfLengths(self, arr: List[int], target: int) -> int:
        n = len(arr)
        dp = [float('inf')] * n
        l, curSum = 0, 0
        ans, curShortest = float('inf'), float('inf')

        for r, num in enumerate(arr):
            curSum += num
            while curSum > target: # shrink window if past target
                curSum -= arr[l]
                l += 1
            
            if curSum == target:
                # dont care about l == 0 because it would get dp[-1] which would be inf
                # and we wouldnt want to update ans to anything meaningful anyways
                ans = min(ans, dp[l - 1] + (r - l + 1))
                curShortest = min(curShortest, r - l + 1)
            
            dp[r] = curShortest
        
        return ans if ans < float('inf') else -1
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