Minimum Size Subarray in Infinite Array
<p>You are given a <strong>0-indexed</strong> array <code>nums</code> and an integer <code>target</code>.</p>

<p>A <strong>0-indexed</strong> array <code>infinite_nums</code> is generated by infinitely appending the elements of <code>nums</code> to itself.</p>

<p>Return <em>the length of the <strong>shortest</strong> subarray of the array </em><code>infinite_nums</code><em> with a sum equal to </em><code>target</code><em>.</em> If there is no such subarray return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3], target = 5
<strong>Output:</strong> 2
<strong>Explanation:</strong> In this example infinite_nums = [1,2,3,1,2,3,1,2,...].
The subarray in the range [1,2], has the sum equal to target = 5 and length = 2.
It can be proven that 2 is the shortest length of a subarray with sum equal to target = 5.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,1,2,3], target = 4
<strong>Output:</strong> 2
<strong>Explanation:</strong> In this example infinite_nums = [1,1,1,2,3,1,1,1,2,3,1,1,...].
The subarray in the range [4,5], has the sum equal to target = 4 and length = 2.
It can be proven that 2 is the shortest length of a subarray with sum equal to target = 4.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,4,6,8], target = 3
<strong>Output:</strong> -1
<strong>Explanation:</strong> In this example infinite_nums = [2,4,6,8,2,4,6,8,...].
It can be proven that there is no subarray with sum equal to target = 3.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= target &lt;= 10<sup>9</sup></code></li>
</ul>
Array; Hash Table; Prefix Sum; Sliding Window
class Solution {
    public int minSizeSubarray(int[] nums, int target) {
        long s = Arrays.stream(nums).sum();
        int n = nums.length;
        int a = 0;
        if (target > s) {
            a = n * (target / (int) s);
            target -= target / s * s;
        }
        if (target == s) {
            return n;
        }
        Map<Long, Integer> pos = new HashMap<>();
        pos.put(0L, -1);
        long pre = 0;
        int b = 1 << 30;
        for (int i = 0; i < n; ++i) {
            pre += nums[i];
            if (pos.containsKey(pre - target)) {
                b = Math.min(b, i - pos.get(pre - target));
            }
            if (pos.containsKey(pre - (s - target))) {
                b = Math.min(b, n - (i - pos.get(pre - (s - target))));
            }
            pos.put(pre, i);
        }
        return b == 1 << 30 ? -1 : a + b;
    }
}