<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 ctOfSub(vector<int>nums,int k){
        int n=nums.size();
        int l=0,r=0,ct=0;
        unordered_map<int,int>mp;
        while(r<n){
             mp[nums[r]%2]+=1;
             while(mp[1]>k){
                mp[nums[l]%2]--;
                l++;
             }
             if(mp[1]<=k) ct+=(r-l+1);
             r++;
        }
        return ct;
    }
    int numberOfSubarrays(vector<int>& nums, int k) {
        int f=ctOfSub(nums,k);
        int s=ctOfSub(nums,k-1);
        cout<<f<<" "<<s<<endl;
        return (f-s);
    }
};
66571