You are given an array A of N integers and an integer S. Your task is to compute how many ways one can choose a contiguous fragment of A that has an arithmetic mean equal to S. The arithmetic mean (average) of a fragment is the sum of the elements of the fragment divided by its length. For example, the arithmetic mean of [1, 3, 3, 5] = 12/4 = 3. Write a function which returns the number of contiguous fragments of A whose arithmetic means are equal to S. Examples: Given A = [3, 2, 4] and S = 3, your function should return 3, since the arithmetic means of fragments [3], [2, 4], [3, 2, 4] are equal to 3. Given A = [-1, 3, 2, -2] and S = 1, your function should return 2, since fragments [-1, 3] and [3, 2, -2] have an arithmetic mean of 1. Given A = [2, 3, 6] and S = 5, your function should return 0, since there exist no contiguous fragments whose arithmetic mean is equal to 5. Write an efficient algorithm for the following assumptions: N is an integer within the range [1, ..., 100,000] S is an integer within the range [-100,000,000,000, 100,000,000,000] Each element of array A is an integer within the range [-100,000,000,000, 100,000,000,000]

# Implementing the function to count the number of contiguous fragments with an arithmetic mean equal to S def count_fragments(A: list, S: int) -> int: count = 0 # Initialize the counter for fragments with mean equal to S n = len(A) # Length of the array A # Two-pointer approach for i in range(n): fragment_sum = 0 # Initialize the sum of the current fragment for j in range(i, n): fragment_sum += A[j] # Update the sum of the current fragment # Calculate the arithmetic mean of the current fragment mean = fragment_sum / (j - i + 1) # If the mean equals S, update the counter if mean == S: count += 1 return count # Test the function print(count_fragments([3, 2, 4], 3)) # Should return 3 print(count_fragments([-1, 3, 2, -2], 1)) # Should return 2 print(count_fragments([2, 3, 6], 5)) # Should return 0

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Arithmetic Mean

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