Find Two Non-overlapping Sub-arrays Each With Target Sum
<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>
Array; Hash Table; Binary Search; Dynamic Programming; Sliding Window
class Solution {
    public int minSumOfLengths(int[] arr, int target) {
        Map<Integer, Integer> d = new HashMap<>();
        d.put(0, 0);
        int n = arr.length;
        int[] f = new int[n + 1];
        final int inf = 1 << 30;
        f[0] = inf;
        int s = 0, ans = inf;
        for (int i = 1; i <= n; ++i) {
            int v = arr[i - 1];
            s += v;
            f[i] = f[i - 1];
            if (d.containsKey(s - target)) {
                int j = d.get(s - target);
                f[i] = Math.min(f[i], i - j);
                ans = Math.min(ans, f[j] + i - j);
            }
            d.put(s, i);
        }
        return ans > n ? -1 : ans;
    }
}