Sum of All Odd Length Subarrays

class Solution {
public:
    int sumOddSubarrays(vector<int>& arr){
        int len = int(arr.size()), total = 0;

        for (int idx = 0; idx < len; ++idx) {
            int left_count = idx, right_count = len - idx - 1;
            total += arr[idx] * (left_count / 2 + 1) * (right_count / 2 + 1);
            total += arr[idx] * ((left_count + 1) / 2) * ((right_count + 1) / 2);
        }

        return total;
    }
};

The solution involves calculating the sum of all odd-length subarrays for a given array. This is achieved by iterating through each element in the array, determining the contribution of each element to all possible odd length subarrays that include it.

Here's a step-by-step breakdown of the approach used in the provided C++ code:

1. Initialize total to zero. This variable will accumulate the sum of all subarrays.

2. Use a for loop to iterate through the array. The loop variable idx represents the current index.

3. For each element at index idx, compute:

   • left_count: the number of elements to the left of the element (inclusive).
   • right_count: the number of elements to the right of the element (inclusive).

4. For each element, calculate its contribution to subarrays where it is the starting or ending element within an odd length subarray. This involves calculating:

   • Multiplying the current element by the number of times it would appear in subarrays formed by elements up to and including the current element (left_count) and elements after the current element including itself (right_count).

   Perform this multiplication twice and sum them:

   • First considering subarrays starting or ending right at the element itself.
   • Second considering subarrays that start or end beyond the current element but still include it.

5. Accumulate these contributions to total.

The result, total, represents the sum of all elements across all odd-length subarrays of the given array. This efficient approach ensures each element's contribution based on its position is calculated directly, thus reducing the need for nested loops explicitly forming each subarray.

class Solution {
    public int calculateOddSum(int[] array) {
        int size = array.length, totalSum = 0;
        
        for (int index = 0; index < size; ++index) {
            int leftCount = index, rightCount = size - index - 1;
            totalSum += array[index] * (leftCount / 2 + 1) * (rightCount / 2 + 1);
            totalSum += array[index] * ((leftCount + 1) / 2) * ((rightCount + 1) / 2);
        }
        
        return totalSum;
    }
}

The code provided efficiently calculates the sum of all subarrays with odd lengths within a given integer array in Java. The calculateOddSum method, encapsulated within the Solution class, leverages a mathematical approach to avoid generating all possible subarrays explicitly, which would be computationally expensive.

Here's how the algorithm operates:

1. Initialize totalSum to 0 to keep track of the cumulative sum of all elements across odd length subarrays.
2. Iterate through each element in the array using a loop where index represents the current position within the array.
3. For each element, calculate leftCount as the number of elements to the left of the current index (inclusive) and rightCount as the number of elements to the right (inclusive).
4. The odd-length subarray contribution of the current element is calculated by taking the product of:
   • The current element (array[index]),
   • The count of ways to form odd-length subarrays including the left elements (derived either by combining an odd or even number on the left with an odd or even on the right respectively to maintain overall odd length).
5. Add the result to totalSum.

This algorithm efficiently sums up contributions from each array element, respecting the constraints of only considering odd-length subarrays. The mathematical insight used here reduces the need for deeper nested loops, therefore optimizing the overall computational time to O(n), where n is the number of elements in the array.

class Solution:
    def calculateOddSum(self, array: List[int]) -> int:
        total_length = len(array)
        total_sum = 0
        
        for index, value in enumerate(array):
            left_count, right_count = index, total_length - index - 1
            total_sum += value * ((left_count // 2) + 1) * ((right_count // 2) + 1)
            total_sum += value * ((left_count + 1) // 2) * ((right_count + 1) // 2)
        return total_sum

The given Python solution defines a method calculateOddSum for calculating the sum of all subarrays with odd lengths within an array. The method uses mathematical reasoning to efficiently compute the result without needing to explicitly generate all subarrays.

• Calculate the length of the input array.
• Initialize a variable to accumulate the total sum of desired subarray values.
• Iterate over each element in the array using its index:
  • Determine how many times the current element contributes to subarrays where the count of elements is odd using combinations of possible starts (left) and ends (right).
  • Adjust the count by considering different possible odd lengths focusing on how the indices divide.
  • Sum up the contributions of the current element to the total, taking into account both possible symmetric distributions from left to right.
• Return the accumulated total sum of all such subarrays.

This method leverages combinatorial logic to count contributions of each array item to the sum based on their positions, thus allowing the sum calculation in linear time relative to the size of the input array.