# 1588. Sum of All Odd Length Subarrays

## Description

Given an array of positive integers arr, return the sum of all possible odd-length subarrays of arr.

A subarray is a contiguous subsequence of the array.

Example 1:

Input: arr = [1,4,2,5,3]
Output: 58
Explanation: The odd-length subarrays of arr and their sums are:
[1] = 1
[4] = 4
[2] = 2
[5] = 5
[3] = 3
[1,4,2] = 7
[4,2,5] = 11
[2,5,3] = 10
[1,4,2,5,3] = 15
If we add all these together we get 1 + 4 + 2 + 5 + 3 + 7 + 11 + 10 + 15 = 58

Example 2:

Input: arr = [1,2]
Output: 3
Explanation: There are only 2 subarrays of odd length, [1] and [2]. Their sum is 3.

Example 3:

Input: arr = [10,11,12]
Output: 66


Constraints:

• 1 <= arr.length <= 100
• 1 <= arr[i] <= 1000

Could you solve this problem in O(n) time complexity?

## Solutions

• class Solution {
public int sumOddLengthSubarrays(int[] arr) {
int n = arr.length;
int ans = 0;
for (int i = 0; i < n; ++i) {
int s = 0;
for (int j = i; j < n; ++j) {
s += arr[j];
if ((j - i + 1) % 2 == 1) {
ans += s;
}
}
}
return ans;
}
}

• class Solution {
public:
int sumOddLengthSubarrays(vector<int>& arr) {
int n = arr.size();
int ans = 0;
for (int i = 0; i < n; ++i) {
int s = 0;
for (int j = i; j < n; ++j) {
s += arr[j];
if ((j - i + 1) & 1) {
ans += s;
}
}
}
return ans;
}
};

• class Solution:
def sumOddLengthSubarrays(self, arr: List[int]) -> int:
ans, n = 0, len(arr)
for i in range(n):
s = 0
for j in range(i, n):
s += arr[j]
if (j - i + 1) & 1:
ans += s
return ans


• func sumOddLengthSubarrays(arr []int) (ans int) {
n := len(arr)
for i := range arr {
s := 0
for j := i; j < n; j++ {
s += arr[j]
if (j-i+1)%2 == 1 {
ans += s
}
}
}
return
}

• function sumOddLengthSubarrays(arr: number[]): number {
const n = arr.length;
let ans = 0;
for (let i = 0; i < n; ++i) {
let s = 0;
for (let j = i; j < n; ++j) {
s += arr[j];
if ((j - i + 1) % 2 === 1) {
ans += s;
}
}
}
return ans;
}


• impl Solution {
pub fn sum_odd_length_subarrays(arr: Vec<i32>) -> i32 {
let n = arr.len();
let mut ans = 0;
for i in 0..n {
let mut s = 0;
for j in i..n {
s += arr[j];
if (j - i + 1) % 2 == 1 {
ans += s;
}
}
}
ans
}
}