Question

1588. Sum of All Odd Length Subarrays

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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

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

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;
  }
};

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
  }
}

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;
}