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The brute force approach involves iterating over every possible subarray, checking if its length is odd, and then summing its elements. This approach is straightforward but not optimal.
Time Complexity: O(n^3) where n is the number of elements in the array, due to triple nested loops.
Space Complexity: O(1), only a few extra variables are utilized.
1public class Main {
2 public static int sumOddLengthSubarrays(int[] arr) {
3 int totalSum = 0;
4 int n = arr.length;
5 for (int start = 0; start < n; ++start) {
6 for (int end = start; end < n; ++end) {
7 if ((end - start + 1) % 2 != 0) {
8 for (int k = start; k <= end; ++k) {
9 totalSum += arr[k];
10 }
11 }
12 }
13 }
14 return totalSum;
15 }
16
17 public static void main(String[] args) {
18 int[] arr = {1, 4, 2, 5, 3};
19 System.out.println(sumOddLengthSubarrays(arr));
20 }
21}In this Java solution, we apply a similar approach of using nested loops to generate possible subarrays and summing those with an odd length. Java's int[] is used to handle the array input.
The optimized approach calculates the contribution of each element to the final sum using a mathematical formula. For each element at index i, calculate how many odd-length subarrays it can contribute to, then sum directly based on these contributions.
Time Complexity: O(n), much improved by calculating contributions directly.
Space Complexity: O(1), only a few extra integer variables.
1
class Program {
static int SumOddLengthSubarrays(int[] arr) {
int totalSum = 0;
int n = arr.Length;
for (int i = 0; i < n; i++) {
int leftCount = i + 1;
int rightCount = n - i;
int oddCount = (leftCount * rightCount + 1) / 2;
totalSum += oddCount * arr[i];
}
return totalSum;
}
static void Main() {
int[] arr = {1, 4, 2, 5, 3};
Console.WriteLine(SumOddLengthSubarrays(arr));
}
}This C# solution similarly relies on counting how each element participates in possible subarrays, thus leveraging mathematical simplification for efficiency.