<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();
       vector<int>odd(n,0),even(n,0),suff(n,0);
       if(nums[0]%2!=0){
        odd[0]=1;
       }
       else if(nums[0]%2==0){
        even[0]=1;
       }
       
       for(int i=1;i<n;i++){
          if(nums[i]%2!=0){
            odd[i]=odd[i-1]+1;
          }
          else{
            odd[i]=odd[i-1];
          }
       } 
       unordered_map<int,int>mp;
       mp[0]++;
       for(int i=0;i<n;i++){
          mp[odd[i]]++;
       }
       int ct=0;
       for(int i=0;i<n;i++){
        if(odd[i]>=k){
            if(mp.find(odd[i]-k)!=mp.end()){
                ct+=mp[odd[i]-k];
            }
        }
       }
       return ct;
    }
};
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