<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) {
        vector<int> oddEvenArr;
        for (int i = 0; i < nums.size(); i++) {
            if(nums[i] & 1) oddEvenArr.push_back(1);
            else oddEvenArr.push_back(0);
        }
        int curSum = 0, size = nums.size(), count = 0;
        unordered_map<int, int> prevSumMap;
        for (int i = 0; i < size; i++) {
            curSum += oddEvenArr[i];
            if (curSum == k)
                count++;
             if (prevSumMap.find(curSum - k) != prevSumMap.end()){
                count += prevSumMap[curSum - k];
            }
            prevSumMap[curSum]++;
        }
        return count;
    }
};
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