Given an array of positive integers arr, calculate the sum of all possible odd-length subarrays.
A subarray is a contiguous subsequence of the array.
Return the sum of all odd-length subarrays of arr.
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
public class Solution {
public int SumOddLengthSubarrays(int[] arr) {
int res = 0;
var len = arr.Length;
for (int i = 0; i < len; ++i) {
res += ((i + 1) * (len - i) + 1) / 2 * arr[i];
}
return res;
}
}
Time Complexity: O(n)
Space Complexity: O(1)
Consider the subarray that contains A[i],
we can take 0,1,2..,i elements on the left,
from A[0] to A[i],
we have i + 1 choices.
we can take 0,1,2..,n-1-i elements on the right,
from A[i] to A[n-1],
we have n - i choices.
In total, there are (i + 1) * (n - i) subarrays, that contains A[i].
And there are ((i + 1) * (n - i) + 1) / 2 subarrays with odd length, that contains A[i].
A[i] will be counted ((i + 1) * (n - i) + 1) / 2 times.
Example of array [1,2,3,4,5]
1 2 3 4 5 subarray length 1
1 2 X X X subarray length 2
X 2 3 X X subarray length 2
X X 3 4 X subarray length 2
X X X 4 5 subarray length 2
1 2 3 X X subarray length 3
X 2 3 4 X subarray length 3
X X 3 4 5 subarray length 3
1 2 3 4 X subarray length 4
X 2 3 4 5 subarray length 4
1 2 3 4 5 subarray length 5
5 8 9 8 5 total times each index was added.
3 4 5 4 3 total times in odd length array with (x + 1) / 2
2 4 4 4 2 total times in even length array with x / 2