Count Number of Nice Subarrays - Solution & Explanation

The main idea is to use a sliding window to keep track of the current subarray and count of odd numbers. If the count of odd numbers becomes more than k, move the start pointer to decrease the count. For every suitable subarray, calculate the number of possible subarrays that end at the current end pointer. This way, we can efficiently count the nice subarrays.

This solution uses a hash map (dictionary) to count the frequency of certain odd counts (prefix counts). As we iterate over nums, we update the current count of odd numbers. If the current count minus k has been seen before, it means there is a subarray ending at the current position with exactly k odd numbers.

This approach involves checking every possible subarray of nums and counting the odd numbers in each. If a subarray contains exactly k odd numbers, it is counted as nice. While straightforward to implement, this method is not efficient for large arrays as its time complexity is quadratic.

This Python solution iterates over all possible subarrays starting from each index. For each subarray, it counts the number of odd numbers and checks if this matches k. If matched, the count of nice subarrays is incremented.