<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>

class Solution {
public:
    int numSubarraysWithSum(vector<int>& nums, int goal) {
        int n = nums.size();
        int res = 0, t = 0;
        if (goal == 0){
            for (int i = 0; i < n; i++){
                if (nums[i] == 0){
                    int j = i; 
                    while (j + 1 < n && nums[j + 1] == 0) j++;
                    res += ((j - i + 1)*(j - i + 2)/2);
                    i = j;
                }
            }
            return res;
        }
        vector<int> conSum(n + 1, 0);
        vector<int> counting(n + 1, 0);
        counting[0] = 1;
        for (int i = 1; i <= n; i++) {
            conSum[i] = nums[i - 1] + conSum[i - 1];
            counting[conSum[i]]++;
            t = conSum[i] - goal;
            if (t >= 0){
                res += counting[t];
            }
        }
        return res;
    }
};
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