Count Number of Nice Subarrays - Two Pointers / Sliding Window

Problems
1

https://algo.monster/problems/count_number_of_nice_subarrays

2

def count_nice_subarrays(k: int, arr: List[int]) → int:

left = 0
result = 0
final_array = []

while left < len(arr) - k + 1:
    right = left
    sub_array = []
    num_odd = k
    while num_odd and right < len(arr):
        sub_array.append(arr[right])
        if arr[right] % 2 != 0:
            num_odd -= 1
        right += 1
        if num_odd == 0 and sub_array:
            result += 1
    final_array.append(sub_array)
    left += 1
print(final_array)
return result
3

Can anyone please share how for the third case, there are 12 sub arrays containing 3 odd numbers…?
There should be only 4
[2, 4, 5, 7, 8, 10, 11], [4, 5, 7, 8, 10, 11], [5, 7, 8, 10, 11], [7, 8, 10, 11, 12, 14, 15]

4

Below are the subarray with odd for the 3rd use case
2 4 5 7 8 10 11
2 4 5 7 8 10 11 12
2 4 5 7 8 10 11 12 14
4 5 7 8 10 11
4 5 7 8 10 11 12
4 5 7 8 10 11 12 14
5 7 8 10 11
5 7 8 10 11 12
5 7 8 10 11 12 14
7 8 10 11 12 14 15
7 8 10 11 12 14 15 18
7 8 10 11 12 14 15 18 20

5

Below are the combinations for use case 3

2 4 5 7 8 10 11
2 4 5 7 8 10 11 12
2 4 5 7 8 10 11 12 14
4 5 7 8 10 11
4 5 7 8 10 11 12
4 5 7 8 10 11 12 14
5 7 8 10 11
5 7 8 10 11 12
5 7 8 10 11 12 14
7 8 10 11 12 14 15
7 8 10 11 12 14 15 18
7 8 10 11 12 14 15 18 20

6

[2, 4, 5, 7, 8, 10, 11], [2, 4, 5, 7, 8, 10, 11, 12], [2, 4, 5, 7, 8, 10, 11, 12, 14], [4, 5, 7, 8, 10, 11], [4, 5, 7, 8, 10, 11, 12], [4, 5, 7, 8, 10, 11, 12, 14], [5, 7, 8, 10, 11], [5, 7, 8, 10, 11, 12], [5, 7, 8, 10, 11, 12, 14], [7, 8, 10, 11, 12, 14, 15], [7, 8, 10, 11, 12, 14, 15, 18], [7, 8, 10, 11, 12, 14, 15, 18, 20].

7

My javascript solution is too slow for the last case. And the solution provided is too mathematically difficult for me to understand, even the graphics didn’t help. I think author should work on the solutions to make it more intuitive to understand given the price of premium is not cheap…

function countNiceSubarrays(k, arr) {
let a = 0
let b = a + k
let ans = 0

function checkOdd(array, k) {
    let count = k
    for(let i = 0; i<array.length; i++){
        if(array[i]%2 !== 0){
            count--
        }
    }
    if(count == 0){
        return true
    } else {
        return false
    }
}

while(a < arr.length - k + 1){
    let subArray = arr.slice(a, b)
    let isNice = checkOdd(subArray, k)

   
    //if b < arr.length && odd number < k
    if(b<arr.length && !isNice){
        b++  
    }
    //if b == arr.length  && odd number < k
    else if( b == arr.length && !isNice){
        ++a
        b = a + k
        if(a == 0 && b == arr.length && !isNice){
            return ans
        }
    } 
    //if b < arr.length  && add number == k
    else if(b<arr.length && isNice){
        b++
        ans++
    } 
    //if b == arr.length  && add number == k
    else if(b == arr.length && isNice){
        ++a 
        b = a + k
        ans++
    }
}


return ans;

}