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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

 

Follow up:

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

Solutions

Solution 1

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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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
    }
}

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int sumOddLengthSubarrays(int* arr, int arrSize) {
    int ans = 0;
    for (int i = 0; i < arrSize; ++i) {
        int s = 0;
        for (int j = i; j < arrSize; ++j) {
            s += arr[j];
            if ((j - i + 1) % 2 == 1) {
                ans += s;
            }
        }
    }
    return ans;
}

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