1248. Count Number of Nice Subarrays
Question
Given an array of integers nums
and an integer k
. A continuous subarray is called nice if there are k
odd numbers on it.
Return the number of nice subarrays.
Example 1:
Input: nums = [1,1,2,1,1], k = 3
Output: 2
Explanation: The only subarrays with 3 odd numbers are [1,1,2,1] and [1,2,1,1].
Example 2:
Input: nums = [2,4,6], k = 1
Output: 0
Explanation: There are no odd numbers in the array.
Example 3:
Input: nums = [2,2,2,1,2,2,1,2,2,2], k = 2
Output: 16
Constraints:
1 <= nums.length <= 50000
1 <= nums[i] <= 10^5
1 <= k <= nums.length
Algorithm

The nice sub array is the subarray with k odd number

I was thinking of using sliding window , why? Because we are manipulating subarrays and sliding window can help in most case. (Note: non negative elements array)

But I was stuck at the shrink > what should I do to record the current accumulated result or find res. For example
[2,2,2,1,2,2,1,2,2,2]
when the left is 1 and right is 1, how we get the current qualified subarray count from this ? 
We can't with existing values > additional information needed > lastPopped
Step 4 is the mind gap with the answer.

With lastPopped, we can get the sub result (subarrays count starting from left, ending at right) > initialGap

The gap is calculated by left  last popped > this is the number / count current subarray can propose . For example
[2,2,2,1,2,2,1,2,2,2]
when the left is 1 and right is 1, the gap is 4 > why 
It's
[2,2,2,1,2,2,1]
,[2,2,1,2,2,1]
,[2,1,2,2,1]
,[1,2,2,1]
, so the gap is the count
Implementation
class Solution { public int numberOfSubarrays(int[] nums, int k) { Queue<Integer> queue = new LinkedList<>(); int res = 0; int lastPopped = 1; int initialGap = 1; for (int i = 0; i < nums.length; i++) { if (nums[i] % 2 == 1) { queue.offer(i); } if (queue.size() > k) { lastPopped = queue.poll(); } if (queue.size() == k) { initialGap = queue.peek()  lastPopped; res += initialGap; } } return res; } }