<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 soln(vector<int> & arr, int k){
        if(k<0) return 0;
        int res=0;
        int sum=0;
        int j=0;
        for(int i=0;i<arr.size();i++){
            sum+=arr[i];
            while(sum>k && j<=i){
                sum-=arr[j];
                j++;
            }
                res +=i-j+1;
            
        }
        return res;
    }
    int numberOfSubarrays(vector<int>& nums, int k) {
        vector<int> arr =nums;
        for(int i=0;i<nums.size();i++){
            if(arr[i]%2 == 0){
                arr[i] = 0;
            }else arr[i] = 1;
        }
        return soln(arr, k) - soln(arr, k-1);
    }
};
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