1524. Number of Sub-arrays With Odd Sum

1524. Number of Sub-arrays With Odd Sum (Medium)

Given an array of integers arr. Return the number of sub-arrays with odd sum.

As the answer may grow large, the answer must be computed modulo 10^9 + 7.

 

Example 1:

Input: arr = [1,3,5]
Output: 4
Explanation: All sub-arrays 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 sub-arrays 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

Example 4:

Input: arr = [100,100,99,99]
Output: 4

Example 5:

Input: arr = [7]
Output: 1

 

Constraints:

• 1 <= arr.length <= 10^5
• 1 <= arr[i] <= 100

Related Topics:
Array, Math

Solution 1.

Assume sum[i] is the sum of A[0..(i - 1)], sum[0] = 0. We can get any subarray sum using sum[j] - sum[i] = A[i] + .. + A[j-1]. Given 0 < j <= N, to find how many 0 <= i < j can form odd subarray sum with j, we just need to know how many sum[i] are odd or even numbers.

If sum[j] is odd, then we use those even sum[i] to form odd subarray sum.

If sum[j] is even, then we use those odd sum[i] to form odd subarray sum.

So we just need to store how many odd sum[i] we've seen thus far in a variable odd. The number of even sum[i] is j + 1 - odd.

// OJ: https://leetcode.com/problems/number-of-sub-arrays-with-odd-sum/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int numOfSubarrays(vector<int>& A) {
        long mod = 1e9+7, ans = 0, odd = 0, sum = 0;
        for (int i = 0; i < A.size(); ++i) {
            sum += A[i];
            if (sum % 2) ans = (ans + i + 1 - odd) % mod; // sum is odd, use those `i + 1 - odd` even numbers to form odd subarray sums.
            else ans = (ans + odd) % mod; // sum is even, use those `odd` odd numbers to form odd subarray sums.
            odd += sum % 2;
        }
        return ans;
    }
};