<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 n = nums.size();
        for(int i=0; i<n; i++)
        {
            nums[i] = nums[i]%2;
        }
        int res = 0;
        int start = 0;
        int curr = 0;
        vector<vector<int>> ind;
        vector<int> p(2);
        for(int end=0; end<n; end++)
        {
            // cout << "end: " << end << endl;
            curr += nums[end];
            // cout << "curr: " << curr << endl;

            while(curr > k && start <= end)
            {
                curr -= nums[start];
                start++;
                // cout << "curr: " << curr << " start: " << start << endl;
                // cout << "cond: " << (curr > k && start <= end) << endl;
            }

            if(curr == k)
            {
                p[0] = start;
                p[1] = end;
                // cout << start << " " << end << endl;
                ind.push_back(p);
            }
        }
        for(int i=0; i<ind.size(); i++)
        {
            int s = ind[i][0], e = ind[i][1];
            res++;
            for(int i=s; i<e; i++)
            {
                if(nums[i] != 1) res++;
                else break; 
            }
        }
        return res;
    }
};
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