<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>

auto init = [](){
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    return 0;
}();

class Solution {
public:
    // Same as 930. Binary Subarrays With Sum with a tweak
    int numberOfSubarrays(vector<int>& nums, int k) {
        return get(nums,k)-get(nums,k-1);
    }

     int get(vector<int>&nums, int goal){
        if(goal<0) return 0;
        int n = nums.size();
        int l = 0, r = 0;
        int cnt = 0;
        int sum = 0;
        while(r<n){
          sum = sum + (nums[r]%2);
          while(sum>goal){
            sum -= (nums[l]%2);
            l++;
            }
            cnt += (r-l+1);
            r++;
        }
        return cnt;
    }
};
16705