To solve the problem, we need to compute \\((x - 14)^{10}\\), where \\(x\\) is the number of vowels in the answer. This is a self-referential problem, so we need to find a value of the answer such that the number of vowels in the answer matches the \\(x\\) used in the expression.

### Step 1: Define the problem
- Let \\(y = (x - 14)^{10}\\), where \\(x\\) is the number of vowels in \\(y\\).
- Vowels are the letters "a," "e," "i," "o," "u" (we'll consider only lowercase for simplicity, unless otherwise specified).
- We need to find a \\(y\\) such that the number of vowels in \\(y\\) equals \\(x\\), and \\(y = (x - 14)^{10}\\).

### Step 2: Trial and error approach
Since \\(x\\) is the number of vowels, and vowels are non-negative integers, \\(x \\geq 0\\). However, for \\((x - 14)^{10}\\) to be meaningful, \\(x\\) should be such that the expression produces a number whose vowel count matches \\(x\\).

Let's try different values of \\(x\\) and compute \\(y\\), then check the number of vowels in \\(y\\).

#### Case 1: \\(x = 16\\)
- \\(y = (16 - 14)^{10} = 2^{10} = 1024\\).
- Write \\(y\\) in words: "one thousand twenty-four."
- Count vowels in "one thousand twenty-four":
  - o (1), e (1), o (2), u (1), a (1), o (2) → 6 vowels.
- \\(x = 16\\), but the number of vowels in \\(y\\) is 6, so \\(16 \\neq 6\\). This doesn't work.

#### Case 2: \\(x = 18\\)
- \\(y = (18 - 14)^{10} = 4^{10}\\).
- \\(4^5 = 1024\\), so \\(4^{10} = (4^5)^2 = 1024^2\\).
- \\(1024^2 = 1,048,576\\).
- Write \\(y\\) in words: "one million forty-eight thousand five hundred seventy-six."
- Count vowels:
  - o (1), e (1), i (1), i (2), o (2), a (1), o (3), u (1), a (2), e (2), u (2), e (3), i (3) → 13 vowels.
- \\(x = 18\\), but the number of vowels in \\(y\\) is 13, so \\(18 \\neq 13\\). This doesn't work.

#### Case 3: \\(x = 15\\)
- \\(y = (15 - 14)^{10} = 1^{10} = 1\\).
- Write \\(y\\) in words: "one."
- Count vowels in "one": **o**n**e** → 2 vowels.
- \\(x = 15\\), but the number of vowels in \\(y\\) is 2, so \\(15 \\neq 2\\). This doesn't work.

#### Case 4: \\(x = 17\\)
- \\(y = (17 - 14)^{10} = 3^{10}\\).
- \\(3^5 = 243\\), so \\(3^{10} = (3^5)^2 = 243^2\\).
- \\(243^2 = 59,049\\).
- Write \\(y\\) in words: "fifty-nine thousand forty-nine."
- Count vowels:
  - i (1), i (2), e (1), o (1), u (1), a (1), i (3), e (2) → 8 vowels.
- \\(x = 17\\), but the number of vowels in \\(y\\) is 8, so \\(17 \\neq 8\\). This doesn't work.

### Step 3: Look for patterns
Let's summarize the results so far:
- \\(x = 14\\), \\(y = 0\\), vowels = 2.
- \\(x = 15\\), \\(y = 1\\), vowels = 2.
- \\(x = 16\\), \\(y = 1024\\), vowels = 6.
- \\(x = 17\\), \\(y = 59,049\\), vowels = 8.
- \\(x = 18\\), \\(y = 1,048,576\\), vowels = 13.

It seems that as \\(x\\) increases, \\(y\\) grows exponentially (since it's \\(x - 14\\) raised to the 10th power), and the number of vowels in \\(y\\) also increases, but not necessarily matching \\(x\\).

### Step 4: Try smaller \\(x\\)
Let's try \\(x\\) closer to 14 but with a focus on smaller vowel counts.

#### Case 5: \\(x = 12\\)
- \\(y = (12 - 14)^{10} = (-2)^{10} = 1024\\).
- Write \\(y\\) in words: "one thousand twenty-four."
- Count vowels: 6 vowels (as calculated earlier).
- \\(x = 12\\), but the number of vowels in \\(y\\) is 6, so \\(12 \\neq 6\\). This doesn't work.

### Step 5: Try to find a fixed point
We need to find an \\(x\\) such that the number of vowels in \\((x - 14)^{10}\\) equals \\(x\\). Let's try \\(x = 2\\) (since we saw small vowel counts earlier).

#### Case 6: \\(x = 2\\)
- \\(y = (2 - 14)^{10} = (-12)^{10}\\).
- \\((-12)^{10} = 12^{10}\\) (since 10 is even).
- \\(12^5 = 248,832\\), so \\(12^{10} = (12^5)^2 = 248,832^2\\).
- \\(248,832^2 = 61,917,364,224\\).
- Write \\(y\\) in words: "sixty-one billion nine hundred seventeen million three hundred sixty-four thousand two hundred twenty-four."
- Count vowels:
  - i (1), o (1), e (1), i (2), i (3), o (2), e (2), e (3), i (4), o (3), o (4), u (1), a (1), o (5), e (4), o (6) → 16 vowels.
- \\(x = 2\\), but the number of vowels in \\(y\\) is 16, so \\(2 \\neq 16\\). This doesn't work.

### Step 6: Conclusion
After trying multiple values of \\(x\\), we observe that the number of vowels in \\(y = (x - 14)^{10}\\) does not consistently match \\(x\\). The problem may not have a solution where the number of vowels in the result exactly equals \\(x\\), or it may require a specific \\(x\\) that we haven't found yet.

Given the complexity and the lack of a clear fixed point, it seems likely that the problem might not have a solution within the constraints, or the solution requires a more advanced approach (e.g., considering different number naming conventions or constraints on \\(x\\)).

If we must provide an answer, we note that no \\(x\\) tried so far satisfies the condition. Further exploration or clarification (e.g., constraints on \\(x\\), naming conventions) may be needed.

You are given two strings s and pattern.

A string x is called almost equal to y if you can change at most one character in x to make it identical to y.

Return the smallest starting index of a substring in s that is almost equal to pattern. If no such index exists, return -1.

A substring is a contiguous non-empty sequence of characters within a string.

Example 1:
Input: s = "abcdefg", pattern = "bcdffg"
Output: 1

Explanation:
The substring s[1..6] == "bcdefg" can be converted to "bcdffg" by changing s[4] to "f".

Example 2:
Input: s = "ababbababa", pattern = "bacaba"
Output: 4

Explanation:
The substring s[4..9] == "bababa" can be converted to "bacaba" by changing s[6] to "c".

Example 3:
Input: s = "abcd", pattern = "dba"
Output: -1

Example 4:
Input: s = "dde", pattern = "d"
Output: 0

Constraints:
1 <= pattern.length < s.length <= 105
s and pattern consist only of lowercase English letters.

class Solution:
    def minStartingIndex(self, s: str, pattern: str) -> int:
        n = len(pattern)

        # Helper function to check if two strings differ by at most one character
        def differsByAtMostOne(s1: str, s2: str) -> bool:
            if len(s1) != len(s2):
                return False

            # Count differences
            diff_count = 0
            for c1, c2 in zip(s1, s2):
                if c1 != c2:
                    diff_count += 1
                if diff_count > 1:
                    return False
            return True

        # Slide window of size n (pattern length) through s
        for i in range(len(s) - n + 1):
            # Get current substring
            substring = s[i:i+n]
            # Check if it differs by at most one character
            if differsByAtMostOne(substring, pattern):
                return i

        return -1

class Solution:
    def minStartingIndex(self, s: str, pattern: str) -> int:
        ans = self.find_almost_equal_substring(s, pattern)
        return ans

    def compute_z(self, s):
        n = len(s)
        Z = [0] * n
        Z[0] = 0  # The first character is the entire string, which is not considered here
        l, r = 0, 0
        for i in range(1, n):
            if i > r:
                l = r = i
                while r < n and s[r - l] == s[r]:
                    r += 1
                Z[i] = r - l
                r -= 1
            else:
                k = i - l
                if Z[k] < r - i + 1:
                    Z[i] = Z[k]
                else:
                    l = i
                    while r < n and s[r - l] == s[r]:
                        r += 1
                    Z[i] = r - l
                    r -= 1
        return Z

    def find_almost_equal_substring(self, s, pattern):
        m = len(pattern)
        n = len(s)
        if m == 0:
            return 0
        if n < m:
            return -1

        # Step 1: Check for exact match
        exact = s.find(pattern)
        if exact != -1:
            return exact

        # Step 2: Compute lcp and lcs arrays using Z-algorithm
        # Compute lcp (longest prefix of pattern starting at s[i])
        concat_lcp = pattern + '#' + s
        Z_lcp = self.compute_z(concat_lcp)
        lcp = [0] * n
        for i in range(n):
            lcp[i] = Z_lcp[m + 1 + i]

        # Compute lcs (longest suffix of pattern ending at s[i])
        reversed_pattern = pattern[::-1]
        reversed_s = s[::-1]
        concat_lcs = reversed_pattern + '#' + reversed_s
        Z_lcs = self.compute_z(concat_lcs)
        lcs = [0] * n
        len_rev_pat = len(reversed_pattern)
        for i in range(n):
            pos_in_rev_s = i
            pos_in_concat = len_rev_pat + 1 + pos_in_rev_s
            if pos_in_concat >= len(Z_lcs):
                lcs_val = 0
            else:
                lcs_val = Z_lcs[pos_in_concat]
            lcs[n - 1 - i] = lcs_val

        min_index = float('inf')

        # Iterate over each possible j (position in pattern)
        for j in range(m):
            required_prefix = j
            required_suffix = m - j - 1

            # Iterate possible starting indices i in s
            for i in range(n - m + 1):
                # Check prefix condition
                if lcp[i] < required_prefix:
                    continue
                # Check suffix condition
                suffix_start = i + j + 1
                if suffix_start >= n:
                    continue
                if lcs[suffix_start] < required_suffix:
                    continue
                # Check if current character is a mismatch
                if s[i + j] != pattern[j]:
                    if i < min_index:
                        min_index = i
                        # Early exit if possible
                        if min_index == 0:
                            return 0
                        # Break the inner loop as we're looking for the earliest i
                        break

        return min_index if min_index != float('inf') else -1