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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

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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
      }
  }
  
  

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