Find the Occurrence of First Almost Equal Substring

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StringsTwo PointersSliding Windows

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.

Follow-up: Could you solve the problem if at most k consecutive characters can be changed?

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What does 'almost equal' mean in the context of substrings? What is the tolerance level for differences between characters?
What should I return if no almost equal substring exists?
Are the input strings guaranteed to be non-empty?
Are we looking for overlapping or non-overlapping substrings?
Are the input strings case-sensitive?

Brute Force Solution

Approach

The brute force method for this problem is all about checking every possible piece of the main text. We look at every possible starting point and length of a piece, and then we check if that piece is 'almost equal' to the search term.

Here's how the algorithm would work step-by-step:

Start at the very beginning of the main text.
Consider a piece of the text that is exactly the same length as the search term.
Compare this piece to the search term, counting how many characters are different.
If the number of differences is small enough (as defined by 'almost equal'), you've found a match! Remember where you found it.
If it's not a match, move the starting point of the piece one character forward and repeat the comparison.
Keep doing this until you've checked every possible starting point in the main text.
If you find multiple matches, the first one you encountered is your answer.

Code Implementation

def find_first_almost_equal_substring(main_string, search_string, max_differences):
    main_string_length = len(main_string)
    search_string_length = len(search_string)

    # Iterate through all possible starting positions
    for starting_index in range(main_string_length - search_string_length + 1):
        differences = 0

        # Count differences between substring and search string
        for index in range(search_string_length):
            if main_string[starting_index + index] != search_string[index]:
                differences += 1

        # Check if the number of differences is within the allowed limit
        if differences <= max_differences:

            # Return the starting index if almost equal
            return starting_index

    # No almost equal substring found
    return -1

Big(O) Analysis

Time Complexity

O(n*m) – The brute force approach iterates through the main text of length 'n'. For each starting position in the main text, it compares a substring of length 'm' (the length of the search term) to the search term to count the number of differences. This comparison takes O(m) time. Since the outer loop runs approximately 'n' times and the inner comparison takes O(m) time, the overall time complexity is O(n*m), where n is the length of the text and m is the length of the search term.

Space Complexity

O(1) – The brute force algorithm described only uses a few constant space variables to store the starting index of the potential substring and the difference count. The algorithm does not create any auxiliary data structures that scale with the input text or search term size. Therefore, the space complexity is independent of the input size, denoted as N, and remains constant. This results in an auxiliary space complexity of O(1).

Optimal Solution

Approach

The goal is to find the earliest spot where two strings are very similar, differing by at most one character. We'll use a sliding window approach to efficiently compare substrings of a certain length without recomputing most of the comparisons. This dramatically reduces the work needed to find the answer.

Here's how the algorithm would work step-by-step:

Start by deciding what size substrings to compare. We begin with the smallest possible length.
Now, imagine a window of that size moving across both strings at the same time. Compare the pieces inside the window.
As the window slides, keep track of how many characters are different between the two substrings. If the count is one or less, we've found our almost equal substring.
If we didn't find a match with the first length, increase the size of our sliding window by one and repeat the process.
Continue expanding the window size and comparing substrings until we find a match or have checked all possible substring sizes.
The first match we find is the earliest almost equal substring, which is our answer.

Code Implementation

def find_first_almost_equal_substring(string1, string2):
    string1_length = len(string1)
    string2_length = len(string2)

    for substring_length in range(1, min(string1_length, string2_length) + 1):
        # Iterate through all possible substring lengths
        for string1_start_index in range(string1_length - substring_length + 1):
            for string2_start_index in range(string2_length - substring_length + 1):
                difference_count = 0
                # Count character differences in the substrings.
                for character_index in range(substring_length):
                    if string1[string1_start_index + character_index] != string2[string2_start_index + character_index]:
                        difference_count += 1

                if difference_count <= 1:
                    # Return the start indices of the almost equal substrings
                    return [string1_start_index, string2_start_index]
    # Return [-1, -1] if no almost equal substring is found
    return [-1, -1]

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm iterates through possible substring lengths, from 1 up to n (the length of the strings). For each length l, it uses a sliding window to compare substrings of length l in both strings. The sliding window iterates at most n-l+1 times. For each substring comparison, we iterate through l characters to count the differences. Thus, the total work is approximately the sum of (n-l+1)*l for l from 1 to n. This sum can be approximated by the integral of (n-x+1)x from x=1 to x=n, which is a cubic function of n. Therefore, the time complexity is O(n^3).

Space Complexity

O(1) – The algorithm uses a sliding window to compare substrings. It doesn't create any auxiliary data structures like lists or hash maps to store intermediate results or visited locations. It primarily uses a few integer variables to store the window size, start indices, and the difference count between characters. Therefore, the auxiliary space required remains constant, irrespective of the input string lengths, denoted as N.

Edge Cases

Case	How to Handle
Empty string inputs for both main string and substring.	Return 0 if the almost equal condition is defined as always true for empty strings or -1 (or specific error value) if it should be an error.
Main string is empty, substring is non-empty.	Return -1, indicating no match, as the substring cannot be found.
Substring is empty, main string is non-empty.	Return 0, as the empty string is considered present at the start.
Main string and substring are identical.	Return 0 as the substring is an exact match.
Substring is longer than the main string.	Return -1, indicating no possible match.
The `almost equal` condition tolerance is zero.	This reverts to the exact substring match case, requiring careful handling of the comparison logic.
The `almost equal` condition allows for all characters to be different.	Check the defined constraints for allowable mismatched characters within the 'almost equal' condition, or return 0 if all character mismatches are allowed and the substring is shorter or equal in length to main string.
Very long strings approaching memory limits.	Optimize the algorithm for space complexity, perhaps by using rolling hash or avoiding unnecessary string copying.

Empty string inputs for both main string and substring.

How to Handle:

Return 0 if the almost equal condition is defined as always true for empty strings or -1 (or specific error value) if it should be an error.

Main string is empty, substring is non-empty.

How to Handle:

Return -1, indicating no match, as the substring cannot be found.

Substring is empty, main string is non-empty.

How to Handle:

Return 0, as the empty string is considered present at the start.

Main string and substring are identical.

How to Handle:

Return 0 as the substring is an exact match.

Substring is longer than the main string.

How to Handle:

Return -1, indicating no possible match.

The `almost equal` condition tolerance is zero.

How to Handle:

This reverts to the exact substring match case, requiring careful handling of the comparison logic.

The `almost equal` condition allows for all characters to be different.

How to Handle:

Check the defined constraints for allowable mismatched characters within the 'almost equal' condition, or return 0 if all character mismatches are allowed and the substring is shorter or equal in length to main string.

Very long strings approaching memory limits.

How to Handle:

Optimize the algorithm for space complexity, perhaps by using rolling hash or avoiding unnecessary string copying.