Count Pairs of Equal Substrings With Minimum Difference

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Strings

You are given two 0-indexed string arrays strs1 and strs2.

A substring of length k with the starting index of i for strs1 is denoted as strs1[i, i+k-1] and represents the substring of strs1 starting from the index i and containing the next k characters.

A pair of strings (x, y) is said to be equal if x == y.

For each k from 1 to min(length(strs1[i]), length(strs2[j])), your task is to:

Find the minimum difference between the starting index i of strs1 and j of strs2, where strs1[i, i+k-1] == strs2[j, j+k-1].
If no such pair i and j exists satisfying the condition strs1[i, i+k-1] == strs2[j, j+k-1], the difference is considered to be -1.

Your task is to return an array diff of size min(length(strs1[i]), length(strs2[j])), where diff[k-1] is the minimum difference calculated for substring length k.

Example 1:

Input: strs1 = ["the","spartan","cat"], strs2 = ["they","are","cat","spartan"]
Output: [4,3,0]
Explanation: 
When k = 1, we have equal substrings "t" in strs1[0] and strs2[0] with a difference of abs(0 - 0) = 0.
We have equal substrings "s" in strs1[1] and strs2[3] with a difference of abs(1 - 3) = 2.
We have equal substrings "c" in strs1[2] and strs2[2] with a difference of abs(2 - 2) = 0.
The minimum difference for k = 1 is 0.
When k = 2, we have equal substrings "th" in strs1[0] and strs2[0] with a difference of abs(0 - 0) = 0.
We have equal substrings "sp" in strs1[1] and strs2[3] with a difference of abs(1 - 3) = 2.
We have equal substrings "ca" in strs1[2] and strs2[2] with a difference of abs(2 - 2) = 0.
The minimum difference for k = 2 is 0.
When k = 3, we have equal substrings "the" in strs1[0] and strs2[0] with a difference of abs(0 - 0) = 0.
We have equal substrings "spa" in strs1[1] and strs2[3] with a difference of abs(1 - 3) = 2.
We have equal substrings "cat" in strs1[2] and strs2[2] with a difference of abs(2 - 2) = 0.
The minimum difference for k = 3 is 0.

Example 2:

Input: strs1 = ["leetcode","loveleetcode","leetcode"], strs2 = ["leetcode","love","leetcode"]
Output: [0,1,0,1,0,2,2,2,2,2]
Explanation: 
When k = 1, we have equal substrings "l" in strs1[0] and strs2[0] with a difference of abs(0 - 0) = 0.
We have equal substrings "l" in strs1[0] and strs2[2] with a difference of abs(0 - 2) = 2.
We have equal substrings "l" in strs1[1] and strs2[0] with a difference of abs(1 - 0) = 1.
We have equal substrings "l" in strs1[1] and strs2[2] with a difference of abs(1 - 2) = 1.
We have equal substrings "l" in strs1[2] and strs2[0] with a difference of abs(2 - 0) = 2.
We have equal substrings "l" in strs1[2] and strs2[2] with a difference of abs(2 - 2) = 0.
The minimum difference for k = 1 is 0.
When k = 2, we have equal substrings "le" in strs1[0] and strs2[0] with a difference of abs(0 - 0) = 0.
We have equal substrings "le" in strs1[0] and strs2[2] with a difference of abs(0 - 2) = 2.
We have equal substrings "le" in strs1[1] and strs2[0] with a difference of abs(1 - 0) = 1.
We have equal substrings "le" in strs1[1] and strs2[2] with a difference of abs(1 - 2) = 1.
We have equal substrings "le" in strs1[2] and strs2[0] with a difference of abs(2 - 0) = 2.
We have equal substrings "le" in strs1[2] and strs2[2] with a difference of abs(2 - 2) = 0.
The minimum difference for k = 2 is 0.
When k = 3, we have equal substrings "lee" in strs1[0] and strs2[0] with a difference of abs(0 - 0) = 0.
We have equal substrings "lee" in strs1[0] and strs2[2] with a difference of abs(0 - 2) = 2.
We have equal substrings "lee" in strs1[1] and strs2[0] with a difference of abs(1 - 0) = 1.
We have equal substrings "lee" in strs1[1] and strs2[2] with a difference of abs(1 - 2) = 1.
We have equal substrings "lee" in strs1[2] and strs2[0] with a difference of abs(2 - 0) = 2.
We have equal substrings "lee" in strs1[2] and strs2[2] with a difference of abs(2 - 2) = 0.
The minimum difference for k = 3 is 0.

Constraints:

1 <= strs1.length, strs2.length <= 50
1 <= strs1[i].length, strs2[i].length <= 50
strs1[i] and strs2[i] consist of only lowercase English letters.

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 is the maximum length of the input string?
Can the input string be empty or null?
If there are multiple pairs of equal substrings with the same minimum difference in length, should I return the count of all such pairs, or is any one such pair sufficient?
By substring, do you mean a contiguous sequence of characters?
If no pairs of equal substrings exist, what should I return? (e.g., 0, -1, null)

Brute Force Solution

Approach

The brute force approach for this problem involves checking all possible pairs of substrings within the given string to find those that are equal. For each potential pair, we calculate the difference in their starting positions. Finally, we identify the pairs with the smallest difference among those that are equal.

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

Consider every possible substring within the string. A substring is just a continuous piece of the original string.
For each substring, compare it to every other possible substring in the string.
If two substrings are equal, meaning they are exactly the same sequence of characters, record the difference between their starting locations.
After comparing all pairs of substrings, find the minimum difference among all the pairs that were equal.
Count the number of pairs that have this minimum difference. These are the pairs of equal substrings with the minimum difference.

Code Implementation

def count_equal_substrings_min_difference(input_string):
    string_length = len(input_string)
    minimum_difference = float('inf')
    equal_pair_count = 0

    for first_substring_start in range(string_length):
        for first_substring_end in range(first_substring_start, string_length):
            first_substring = input_string[first_substring_start:first_substring_end + 1]

            for second_substring_start in range(string_length):
                for second_substring_end in range(second_substring_start, string_length):
                    second_substring = input_string[second_substring_start:second_substring_end + 1]

                    # Compare the two substrings
                    if first_substring == second_substring:

                        difference = abs(first_substring_start - second_substring_start)

                        # Update minimum difference and count
                        if difference < minimum_difference:
                            minimum_difference = difference
                            equal_pair_count = 1

                        elif difference == minimum_difference:
                            # Count pairs with the minimum difference
                            equal_pair_count += 1

    # Handle case where no equal substrings are found
    if minimum_difference == float('inf'):
        return 0

    return equal_pair_count

Big(O) Analysis

Time Complexity

O(n^6) – The algorithm considers all possible substrings, which involves two nested loops, each iterating up to n (the length of the string). Thus, generating all substrings takes O(n^2) time. For each pair of substrings (of which there are O(n^4) pairs), we compare them character by character, taking O(n) time in the worst case. Therefore, the overall time complexity is O(n^2) * O(n^2) * O(n), simplifying to O(n^6).

Space Complexity

O(1) – The described brute force approach does not explicitly use any auxiliary data structures that scale with the input size N (length of the string). It iterates through substrings and compares them, but it doesn't store all substrings in memory. Only a few variables, such as indices and the minimum difference found so far, are used, which require constant space, independent of N.

Optimal Solution

Approach

The goal is to quickly find two equal pieces of text with the smallest possible difference in where they start. We can achieve this by organizing the text in a clever way to avoid unnecessary comparisons.

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

Create a way to group equal pieces of text together based on their content. Imagine you are building a dictionary, where each word is associated with where it appears in the original text.
Go through this dictionary. For each word, look at the places where it appears.
Calculate the difference between each pair of locations for the same word. For example, if the word 'hello' appears at positions 3, 7, and 12, calculate the differences 7-3, 12-7, and 12-3.
Find the smallest difference among all those pairs, that will be our answer.

Code Implementation

def find_min_difference(input_string):
    substring_positions = {}
    minimum_difference = float('inf')

    # Group equal substrings together.
    for index in range(len(input_string)):
        for length in range(1, len(input_string) - index + 1):
            substring = input_string[index:index + length]

            if substring not in substring_positions:
                substring_positions[substring] = []
            substring_positions[substring].append(index)

    for substring, positions in substring_positions.items():
        # Only check substrings that appear more than once.
        if len(positions) > 1:

            positions.sort()
            # Find the minimum difference between positions.
            for i in range(1, len(positions)):
                difference = positions[i] - positions[i - 1]
                minimum_difference = min(minimum_difference, difference)

    if minimum_difference == float('inf'):
        return -1
    else:
        return minimum_difference

Big(O) Analysis

Time Complexity

O(n^2) – Creating the dictionary of substrings to their indices takes O(n) time, assuming a reasonable hash function. However, the crucial part is calculating the differences between all pairs of indices for each substring. In the worst case, a single substring might appear at O(n) positions within the original string. Finding all pairs of those positions requires comparing each position with every other position, leading to O(n^2) comparisons in the worst case for that specific substring. Therefore, the overall time complexity is dominated by this pairwise comparison, resulting in O(n^2).

Space Complexity

O(N) – The algorithm uses a dictionary to store equal pieces of text and their locations. In the worst-case scenario, all substrings of the input string are unique and stored as keys in the dictionary, and each key stores a list of indices, which could be of size N. Therefore, the dictionary could potentially store N keys, each with at most N indices, leading to O(N) space. Other variables like the minimum difference have constant space and are dominated by the dictionary.

Edge Cases

Case	How to Handle
Input array is null or empty	Return 0 or throw an IllegalArgumentException as no substrings can be formed.
Input array contains only one string	Return 0 as we need at least two strings to form a pair.
All strings in the input array are identical	The minimum difference will be 0, and all possible pairs of indices should be counted.
Input array contains duplicate strings at different indices	The solution should correctly identify and count all valid pairs even with duplicates.
No pairs of equal substrings exist	The function should return 0, indicating no such pairs were found.
Strings are very long (approaching maximum string length)	Consider potential memory usage implications when storing or comparing very long substrings.
Integer overflow when calculating the difference between ASCII sums of very long strings.	Use a larger data type (e.g., long) to store the difference of the sums to avoid potential integer overflow issues.
The same substring appears multiple times at different indices	All pairs of indices that satisfy the minimum difference criteria should be counted, avoiding double-counting or omissions.

Input array is null or empty

How to Handle:

Return 0 or throw an IllegalArgumentException as no substrings can be formed.

Input array contains only one string

How to Handle:

Return 0 as we need at least two strings to form a pair.

All strings in the input array are identical

How to Handle:

The minimum difference will be 0, and all possible pairs of indices should be counted.

Input array contains duplicate strings at different indices

How to Handle:

The solution should correctly identify and count all valid pairs even with duplicates.

No pairs of equal substrings exist

How to Handle:

The function should return 0, indicating no such pairs were found.

Strings are very long (approaching maximum string length)

How to Handle:

Consider potential memory usage implications when storing or comparing very long substrings.

Integer overflow when calculating the difference between ASCII sums of very long strings.

How to Handle:

Use a larger data type (e.g., long) to store the difference of the sums to avoid potential integer overflow issues.

The same substring appears multiple times at different indices

How to Handle:

All pairs of indices that satisfy the minimum difference criteria should be counted, avoiding double-counting or omissions.