Binary Subarrays With Sum
<p>Given a binary array <code>nums</code> and an integer <code>goal</code>, return <em>the number of non-empty <strong>subarrays</strong> with a sum</em> <code>goal</code>.</p>

<p>A <strong>subarray</strong> is a contiguous part of the array.</p>

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

<pre>
<strong>Input:</strong> nums = [1,0,1,0,1], goal = 2
<strong>Output:</strong> 4
<strong>Explanation:</strong> The 4 subarrays are bolded and underlined below:
[<u><strong>1,0,1</strong></u>,0,1]
[<u><strong>1,0,1,0</strong></u>,1]
[1,<u><strong>0,1,0,1</strong></u>]
[1,0,<u><strong>1,0,1</strong></u>]
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,0,0,0,0], goal = 0
<strong>Output:</strong> 15
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>nums[i]</code> is either <code>0</code> or <code>1</code>.</li>
	<li><code>0 &lt;= goal &lt;= nums.length</code></li>
</ul>
Array; Hash Table; Prefix Sum; Sliding Window
class Solution {
    public int numSubarraysWithSum(int[] nums, int goal) {
        int[] cnt = new int[nums.length + 1];
        cnt[0] = 1;
        int ans = 0, s = 0;
        for (int v : nums) {
            s += v;
            if (s - goal >= 0) {
                ans += cnt[s - goal];
            }
            ++cnt[s];
        }
        return ans;
    }
}