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Formatted question description: https://leetcode.ca/all/2495.html

# 2495. Number of Subarrays Having Even Product

## Description

Given a 0-indexed integer array nums, return the number of subarrays of nums having an even product.

Example 1:

Input: nums = [9,6,7,13]
Output: 6
Explanation: There are 6 subarrays with an even product:
- nums[0..1] = 9 * 6 = 54.
- nums[0..2] = 9 * 6 * 7 = 378.
- nums[0..3] = 9 * 6 * 7 * 13 = 4914.
- nums[1..1] = 6.
- nums[1..2] = 6 * 7 = 42.
- nums[1..3] = 6 * 7 * 13 = 546.


Example 2:

Input: nums = [7,3,5]
Output: 0
Explanation: There are no subarrays with an even product.


Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 105

## Solutions

• class Solution {
public long evenProduct(int[] nums) {
long ans = 0;
int last = -1;
for (int i = 0; i < nums.length; ++i) {
if (nums[i] % 2 == 0) {
last = i;
}
ans += last + 1;
}
return ans;
}
}

• class Solution {
public:
long long evenProduct(vector<int>& nums) {
long long ans = 0;
int last = -1;
for (int i = 0; i < nums.size(); ++i) {
if (nums[i] % 2 == 0) {
last = i;
}
ans += last + 1;
}
return ans;
}
};

• class Solution:
def evenProduct(self, nums: List[int]) -> int:
ans, last = 0, -1
for i, v in enumerate(nums):
if v % 2 == 0:
last = i
ans += last + 1
return ans


• func evenProduct(nums []int) int64 {
ans, last := 0, -1
for i, v := range nums {
if v%2 == 0 {
last = i
}
ans += last + 1
}
return int64(ans)
}