<p>Given an array of integers <code>nums</code> and an integer <code>k</code>. A continuous subarray is called <strong>nice</strong> if there are <code>k</code> odd numbers on it.</p>

<p>Return <em>the number of <strong>nice</strong> sub-arrays</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,1,2,1,1], k = 3
<strong>Output:</strong> 2
<strong>Explanation:</strong> The only sub-arrays with 3 odd numbers are [1,1,2,1] and [1,2,1,1].
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,4,6], k = 1
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no odd numbers in the array.
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,2,2,1,2,2,1,2,2,2], k = 2
<strong>Output:</strong> 16
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 50000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10^5</code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>

class Solution {
    int atMostK(vector<int>& nums, int k) {
        int start = 0, end = 0;
        int count = 0;
        int odds = 0;

        while (end < nums.size()) {
            if (nums[end] % 2 == 1) {
                odds++;
            }

            while (odds > k) {
                if (nums[start] % 2 == 1)
                    odds--;
                start++;
            }

            // All subarrays ending at `end` and starting from `start` to `end`
            // are valid
            count += end - start + 1;

            end++;
        }

        return count;
    }

public:
    int numberOfSubarrays(vector<int>& nums, int k) {
        return atMostK(nums, k) - atMostK(nums, k - 1);
    }
};
13056