<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 {
public:
    int numberOfSubarrays(vector<int>& nums, int k) {
        int l = 0, r = 0;
        int cnt = 0, oddcnt = 0;
        int prefixCount = 0;

        while (r < nums.size()) {
        // Expand the window by adding the right element
            if (nums[r] % 2 == 1) oddcnt++;

            // Shrink the window if the number of odd numbers exceeds k
            while (oddcnt > k) {
                if (nums[l] % 2 == 1) oddcnt--;
                l++;
            }

            // Count subarrays with exactly k odd numbers
            if (oddcnt == k) {
                int tempL = l;
                while (tempL <= r && oddcnt == k) {
                    if (nums[tempL] % 2 == 1) oddcnt--;
                    tempL++;
                }
                cnt += (tempL - l);
                oddcnt = k; // Restore odd count for the next iteration
            }

            r++;
        }

        return cnt;
    }
};
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