# 1524. Number of Sub-arrays With Odd Sum

## Description

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

## Solutions

• class Solution {
public int numOfSubarrays(int[] arr) {
final int mod = (int) 1e9 + 7;
int[] cnt = {1, 0};
int ans = 0, s = 0;
for (int x : arr) {
s += x;
ans = (ans + cnt[s & 1 ^ 1]) % mod;
++cnt[s & 1];
}
return ans;
}
}

• class Solution {
public:
int numOfSubarrays(vector<int>& arr) {
const int mod = 1e9 + 7;
int cnt[2] = {1, 0};
int ans = 0, s = 0;
for (int x : arr) {
s += x;
ans = (ans + cnt[s & 1 ^ 1]) % mod;
++cnt[s & 1];
}
return ans;
}
};

• class Solution:
def numOfSubarrays(self, arr: List[int]) -> int:
mod = 10**9 + 7
cnt = [1, 0]
ans = s = 0
for x in arr:
s += x
ans = (ans + cnt[s & 1 ^ 1]) % mod
cnt[s & 1] += 1
return ans


• func numOfSubarrays(arr []int) (ans int) {
const mod int = 1e9 + 7
cnt := [2]int{1, 0}
s := 0
for _, x := range arr {
s += x
ans = (ans + cnt[s&1^1]) % mod
cnt[s&1]++
}
return
}

• function numOfSubarrays(arr: number[]): number {
let ans = 0;
let s = 0;
const cnt: number[] = [1, 0];
const mod = 1e9 + 7;
for (const x of arr) {
s += x;
ans = (ans + cnt[(s & 1) ^ 1]) % mod;
cnt[s & 1]++;
}
return ans;
}