<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>& arr, int k) {
        int n = arr.size();

        // Convert array elements to 0 if even, 1 if odd
        for(int i = 0; i < n; i++) {
            arr[i] = arr[i] % 2 == 0 ? 0 : 1;
        }

        // Initialize prefix sum array
        vector<int> prefix(n + 1, 0);
        for(int i = 1; i <= n; i++) {
            prefix[i] = prefix[i - 1] + arr[i - 1];
        }

        // Initialize hash map for prefix sums
        vector<int> hash(50001, 0); // Ensure the size is enough to hold prefix sums
        for(int i = 0; i <= n; i++) {
            hash[prefix[i]] += 1;
        }

        // Traverse the prefix sum array to calculate the number of subarrays
        int sum = 0;
        for(int j = 0; j <= n; j++) {
            if(prefix[j] >= k) {
                sum += hash[prefix[j] - k];
            }
        }
        
        return sum;
    }
};

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