<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) {
        vector<int>vis;
        for(int i=0;i<nums.size();i++){
            if (nums[i]%2==0){
                vis.push_back(0);

            }
            else vis.push_back(1);
        }
        int l=0;
        int r=0;
        int cnt=0;
        int sum =0;
        while(r<nums.size()){
            sum=sum+vis[r];
            while(sum>k){
                sum=sum-vis[l];
                l++;
            }
            cnt=cnt+(r-l+1);
            r++;
            
        }
        r=0;
        l=0;
        int cnt0=0;
        sum=0;
        if (k<0) cnt0=0;
        else {
            while(r<nums.size()){
                sum=sum+vis[r];
                while(sum>(k-1)){
                    sum=sum-vis[l];
                    l++;
                }
                cnt0=cnt0+(r-l+1);
                r++;
            }
        
        }
        return (cnt -cnt0);
        
    }
};
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