Maximum Number of Non-Overlapping Subarrays With Sum Equals Target
<p>Given an array <code>nums</code> and an integer <code>target</code>, return <em>the maximum number of <strong>non-empty</strong> <strong>non-overlapping</strong> subarrays such that the sum of values in each subarray is equal to</em> <code>target</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,1,1,1,1], target = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are 2 non-overlapping subarrays [<strong>1,1</strong>,1,<strong>1,1</strong>] with sum equals to target(2).
</pre>

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

<pre>
<strong>Input:</strong> nums = [-1,3,5,1,4,2,-9], target = 6
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are 3 subarrays with sum equal to 6.
([5,1], [4,2], [3,5,1,4,2,-9]) but only the first 2 are non-overlapping.
</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>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= target &lt;= 10<sup>6</sup></code></li>
</ul>
Greedy; Array; Hash Table; Prefix Sum
func maxNonOverlapping(nums []int, target int) (ans int) {
	n := len(nums)
	for i := 0; i < n; i++ {
		s := 0
		vis := map[int]bool{0: true}
		for ; i < n; i++ {
			s += nums[i]
			if vis[s-target] {
				ans++
				break
			}
			vis[s] = true
		}
	}
	return
}