Sum of Subarray Ranges

class Solution {
public:
    long long rangeDifference(vector<int>& elements) {
        int length = elements.size();
        long long result = 0;
        stack<int> stackHelper;
    
        // Summing all minimum elements ranges
        for (int i = 0; i <= length; ++i) {
            while (!stackHelper.empty() &&
                   (i == length || elements[stackHelper.top()] >= elements[i])) {
                int position = stackHelper.top();
                stackHelper.pop();
                int leftIndex = stackHelper.empty() ? -1 : stackHelper.top();
                result -= (long long)elements[position] * (i - position) * (position - leftIndex);
            }
            stackHelper.push(i);
        }
    
        // Summing all maximum elements ranges
        stackHelper.pop();
        for (int i = 0; i <= length; ++i) {
            while (!stackHelper.empty() &&
                   (i == length || elements[stackHelper.top()] <= elements[i])) {
                int position = stackHelper.top();
                stackHelper.pop();
                int leftIndex = stackHelper.empty() ? -1 : stackHelper.top();
                result += (long long)elements[position] * (i - position) * (position - leftIndex);
            }
            stackHelper.push(i);
        }
        return result;
    }
};

The solution provided involves computing the sum of subarray ranges using a monotonic stack method in C++. The function, named rangeDifference, calculates the difference between the sum of maximum and minimum elements' contributions across all subarrays of a given input vector elements.

Follow these steps to understand the implementation of this solution:

1. Initialize necessary variables:

   • A long long type variable named result to store the calculated range difference.
   • An integer length to hold the size of the input vector.
   • A stack of integers stackHelper to assist with the identification of minimum and maximum elements in subarrays.

2. Calculate contributions of minimum range values in all subarrays:

   • Iterate through each element in the vector with an extra loop iteration to handle edge cases.
   • Use the stack to keep track of indices with increasing values and simultaneously calculate the minimum range sum.
   • For each element, determine the appropriate subarray boundaries and use these to decrease the result.

3. Clear the stack after processing minimum elements to reuse it for maximum elements.

4. Calculate contributions of maximum range values in all subarrays:

   • Similar to the above, but track indices with decreasing values to appropriately calculate the maximum range sum.
   • For each element, determine subarray boundaries and update the result by adding to it.

5. Finally, return the value of result, which now contains the range difference between the sum of all maximum and minimum subarray elements.

This approach efficiently computes the result by leveraging stack operations to keep track of subarray boundaries, avoiding the need for nested loops typically used in brute-force solutions. The complexity remains linear relative to the number of elements in the input vector due to the stack-based implementation.

class Solution {
    
    public long rangeDifference(int[] elements) {
        int length = elements.length;
        long result = 0;
        Stack<Integer> stack = new Stack<>();
    
        // Handle minimum values sum
        for (int end = 0; end <= length; ++end) {
            while (
                !stack.isEmpty() &&
                (end == length || elements[stack.peek()] >= elements[end])
            ) {
                int center = stack.peek();
                stack.pop();
                int start = stack.isEmpty() ? -1 : stack.peek();
                result -= (long) elements[center] * (end - center) * (center - start);
            }
            stack.push(end);
        }
    
        // Handle maximum values sum
        stack.clear();
        for (int end = 0; end <= length; ++end) {
            while (
                !stack.isEmpty() &&
                (end == length || elements[stack.peek()] <= elements[end])
            ) {
                int center = stack.peek();
                stack.pop();
                int start = stack.isEmpty() ? -1 : stack.peek();
                result += (long) elements[center] * (end - center) * (center - start);
            }
            stack.push(end);
        }
        return result;
    }
}

This Java solution tackles the problem of calculating the sum of subarray ranges using a rangeDifference method. The method computes the difference between the sums of maximum values and minimum values of all subarrays within the provided array of integers.

Here's how the code operates:

1. The method rangeDifference accepts an array elements and initializes variables including length for the array length, and result initially set to zero for storing the final difference in sums.
2. A Stack<Integer> is used to manage indices during the computation.

The process is divided into two main phases:

1. Handling minimum values sum:

   • Iterate through the array using a variable end from 0 to length.
   • For each element, if the current stack is not empty and the element at the index of the stack's top is greater than or equal to the current element (or if end equals length), compute the contribution of the subarray with respect to the minimum value and subtract it from result.
   • Push the current index end onto the stack after processing.

2. Handling maximum values sum:

   • Clear the stack for a fresh start.
   • Iterate through the array in a similar manner to the minimum values computation.
   • This time, add the contribution of the subarray with respect to the maximum value to result.

Finally, the method returns result, which now holds the total difference between the sum of maximum values and the sum of minimum values across all subarrays.

This approach efficiently computes the desired range difference utilizing stack data structure to maintain the indices, optimizing the process of subarray range calculation.

class Solution:
    def sumOfSubarrayRanges(self, values: List[int]) -> int:
        length, total = len(values), 0 
        temp_stack = []
            
        # Calculate sum of minimums in all subarrays
        for end_idx in range(length + 1):
            while temp_stack and (end_idx == length or values[temp_stack[-1]] > values[end_idx]):
                middle = temp_stack.pop()
                start = -1 if not temp_stack else temp_stack[-1]
                total -= values[middle] * (middle - start) * (end_idx - middle)
            temp_stack.append(end_idx)
    
        # Calculate sum of maximums in all subarrays
        temp_stack.clear()
        for end_idx in range(length + 1):
            while temp_stack and (end_idx == length or values[temp_stack[-1]] < values[end_idx]):
                middle = temp_stack.pop()
                start = -1 if not temp_stack else temp_stack[-1]
                total += values[middle] * (middle - start) * (end_idx - middle)
            temp_stack.append(end_idx)
            
        return total

Below is a summary of how the given Python code calculates the sum of ranges in subarrays:

• First, define a function within a Solution class called sumOfSubarrayRanges which accepts a list of integers (values).

• Initialize variables for the length of the list (length), the resultant total sum (total), and a temporary stack (temp_stack) to facilitate range calculations.

• Solve for the sum total of minimum values across all the subarrays:

  • Iterate over the range 0 to length + 1.
  • In each iteration, pop elements from temp_stack if it's non-empty and the next value is less than or they're at the last index. Calculate the subarray's range minimum contribution to the total and subtract it.
  • Append the current index to temp_stack whether or not elements were popped.

• Reset temp_stack for second iteration.

• Repeat a similar procedure to calculate the sum of the maximum values in the subarrays:

  • Iterate similarly from 0 to length + 1.
  • Conditions for popping from the stack change to check for larger values, and the contributions are added to the total instead of subtracted.

• The final value of total represents the sum of the ranges of all possible subarrays from the values list, obtained by subtracting all subarray minimum values and adding all subarray maximum values. The function ultimately returns this total.