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
class Solution:
def sumOddLengthSubarrays(self, arr: List[int]) -> int:
n = len(arr)
presum = [0] * (n + 1)
for i in range(n):
presum[i + 1] = presum[i] + arr[i]
res = 0
for i in range(n):
for j in range(0, n, 2):
if i + j < n:
res += presum[i + j + 1] - presum[i]
return res
class Solution {
public int sumOddLengthSubarrays(int[] arr) {
int n = arr.length;
int[] presum = new int[n + 1];
for (int i = 0; i < n; ++i) {
presum[i + 1] = presum[i] + arr[i];
}
int res = 0;
for (int i = 0; i < n; ++i) {
for (int j = 0; i + j < n; j += 2) {
res += presum[i + j + 1] - presum[i];
}
}
return res;
}
}
class Solution {
public:
int sumOddLengthSubarrays(vector<int>& arr) {
int n = arr.size();
int presum[n + 1];
for (int i = 0; i < n; ++i) presum[i + 1] = presum[i] + arr[i];
int res = 0;
for (int i = 0; i < n; ++i)
{
for (int j = 0; i + j < n; j += 2)
{
res += presum[i + j + 1] - presum[i];
}
}
return res;
}
};
func sumOddLengthSubarrays(arr []int) int {
n := len(arr)
presum := make([]int, n+1)
for i := range arr {
presum[i+1] = presum[i] + arr[i]
}
res := 0
for i := 0; i < n; i++ {
for j := 0; i+j < n; j += 2 {
res += presum[i+j+1] - presum[i]
}
}
return res
}