# Solving Leetcode Interviews in Seconds with AI: Number of Sub-arrays With Odd Sum


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1524" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given an array of integers arr, return the number of subarrays with an odd sum. Since the answer can be very large, return it modulo 109 + 7.   Example 1:  Input: arr = [1,3,5] Output: 4 Explanation: All subarrays are [[1],[1,3],[1,3,5],[3],[3,5],[5]] All sub-arrays sum are [1,4,9,3,8,5]. Odd sums are [1,9,3,5] so the answer is 4.  Example 2:  Input: arr = [2,4,6] Output: 0 Explanation: All subarrays are [[2],[2,4],[2,4,6],[4],[4,6],[6]] All sub-arrays sum are [2,6,12,4,10,6]. All sub-arrays have even sum and the answer is 0.  Example 3:  Input: arr = [1,2,3,4,5,6,7] Output: 16    Constraints:  1 <= arr.length <= 105 1 <= arr[i] <= 100  

	# Explanation
	Here's a breakdown of the solution:

*   **Key Idea:** Instead of calculating the sum of every subarray, we can maintain counts of subarrays ending at the current index with even and odd sums.
*   **Dynamic Programming:** We iterate through the array, updating the even and odd counts based on whether the current element is even or odd.
*   **Modulo Arithmetic:**  We apply the modulo operator at each step to prevent integer overflow.

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

	
	# Code
	```python
	def numOfSubarrays(arr):
    """
    Given an array of integers arr, return the number of subarrays with an odd sum.
    Since the answer can be very large, return it modulo 109 + 7.
    """
    MOD = 10**9 + 7
    even_count = 0
    odd_count = 0
    result = 0

    for num in arr:
        if num % 2 == 0:
            even_count += 1
        else:
            new_even_count = odd_count
            new_odd_count = even_count + 1
            even_count = new_even_count
            odd_count = new_odd_count

        result = (result + odd_count) % MOD

    return result
	```