README_EN.md

1524. Number of Sub-arrays With Odd Sum

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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

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

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