<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 answer=0;
        int n=nums.size();
        for(int i=0;i<n;i++){
            int count=0;
            for(int j=i;j<n;j++){
                if(nums[j]%2==1){
                    if(count>k){
                        break;
                    }
                    else{
                        count++;
                        if(count==k){
                            answer++;
                        }
                    }
                
                }
            }
        }
        return answer;
    } */

    int helperfn(vector<int>nums,int k){
    int l=0;
    int r=0;
    int sum=0;
    int answer=0;
    int n=nums.size();
    if(k<0){
        return 0;
    }
    while(r<n){
        sum+=nums[r]%2;
        while(sum>k){
            sum -=nums[l]%2;
            l++;
        }
        answer=answer+(r-l+1);
        r++;
    }
    return answer;
    }

    int numberOfSubarrays(vector<int>& nums, int k) {
     return helperfn(nums,k)- helperfn(nums,k-1);  
    }    
};
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