Problem URL: Number of sub arrays with odd sum

We will start counting both the odd and even numbers in the array. If the current number is even, then the number of odd subarray will the count of odd numbers. Otherwise, we will change the odd by adding 1 to the even numbers, and swap even with the old odd numbers. Then we add odd number to the result on each iteration. Finally, we will return the result by modding with the given mod value.

class Solution:
    def numOfSubarrays(self, arr: List[int]) -> int:
        MOD = 10**9+7

        even, odd = 0, 0
        res = 0

        for x in arr:
            if x % 2 == 0:
                even += 1
            else:
                old_odd = odd
                odd = even + 1
                even = old_odd

            res += odd

        return res % MOD

Time Complexity: O(n)
Space Complexity: O(1)