Approach: Contribution / brute force

def sumOddLengthSubarrays(arr):
    n = len(arr); total = 0
    for i in range(n):
        s = 0
        for j in range(i, n):
            s += arr[j]
            if (j - i + 1) % 2 == 1: total += s
    return total

Complexity: O(n^2) time, O(1) space.

Full Solution — all languages

class Solution:
    def sumOddLengthSubarrays(self, nums):
        n = len(nums)
        total = 0
        for i in range(n):
            s = 0
            for j in range(i, n):
                s += nums[j]
                if (j - i + 1) % 2 == 1:
                    total += s
        return total

import java.util.*;
class Solution {
    public int sumOddLengthSubarrays(int[] nums) {
        int n = nums.length, total = 0;
        for (int i = 0; i < n; i++) {
            int s = 0;
            for (int j = i; j < n; j++) {
                s += nums[j];
                if ((j - i + 1) % 2 == 1) total += s;
            }
        }
        return total;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int sumOddLengthSubarrays(vector<int>& nums) {
        int n = nums.size(), total = 0;
        for (int i = 0; i < n; i++) {
            int s = 0;
            for (int j = i; j < n; j++) {
                s += nums[j];
                if ((j - i + 1) % 2 == 1) total += s;
            }
        }
        return total;
    }
};

/**
 * @param {number[]} nums
 * @return {number}
 */ var sumOddLengthSubarrays = function(nums) {
    const n = nums.length;
    let total = 0;
    for (let i = 0; i < n; i++) {
        let s = 0;
        for (let j = i; j < n; j++) {
            s += nums[j];
            if ((j - i + 1) % 2 === 1) total += s;
        }
    }
    return total;
};

Sum of All Odd Length Subarrays

Example 1

Example 2

Constraints