Smallest Substring With Identical Characters II

Hard

Salesforce

1 view

Topics:

StringsSliding WindowsTwo Pointers

You are given a binary string s of length n and an integer numOps.

You are allowed to perform the following operation on s at most numOps times:

Select any index i (where 0 <= i < n) and flip s[i]. If s[i] == '1', change s[i] to '0' and vice versa.

You need to minimize the length of the longest substring of s such that all the characters in the substring are identical.

Return the minimum length after the operations.

Example 1:

Input: s = "000001", numOps = 1

Output: 2

Explanation:

By changing s[2] to '1', s becomes "001001". The longest substrings with identical characters are s[0..1] and s[3..4].

Example 2:

Input: s = "0000", numOps = 2

Output: 1

Explanation:

By changing s[0] and s[2] to '1', s becomes "1010".

Example 3:

Input: s = "0101", numOps = 0

Output: 1

Constraints:

1 <= n == s.length <= 10⁵
s consists only of '0' and '1'.
0 <= numOps <= n

Try coding it in LeetCode

Smallest Substring With Identical Characters II

Hard

Salesforce

1 view

Topics:

StringsSliding WindowsTwo Pointers

You are given a binary string s of length n and an integer numOps.

You are allowed to perform the following operation on s at most numOps times:

Select any index i (where 0 <= i < n) and flip s[i]. If s[i] == '1', change s[i] to '0' and vice versa.

You need to minimize the length of the longest substring of s such that all the characters in the substring are identical.

Return the minimum length after the operations.

Example 1:

Input: s = "000001", numOps = 1

Output: 2

Explanation:

By changing s[2] to '1', s becomes "001001". The longest substrings with identical characters are s[0..1] and s[3..4].

Example 2:

Input: s = "0000", numOps = 2

Output: 1

Explanation:

By changing s[0] and s[2] to '1', s becomes "1010".

Example 3:

Input: s = "0101", numOps = 0

Output: 1

Constraints:

1 <= n == s.length <= 10⁵
s consists only of '0' and '1'.
0 <= numOps <= n

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 `s`?
Does the string `s` only contain lowercase English letters, or can it contain other characters?
If there are multiple substrings of the same minimum length that satisfy the condition, can I return the length of any of them?
If a character appears more than two times, does that change the expected behavior or output?
Is an empty string a valid input? If so, what should the output be?

Brute Force Solution

Approach

The brute force approach for finding the smallest substring with identical characters involves checking every possible substring. We'll look at substrings of length one, then length two, and so on, until we find one that satisfies our criteria. We essentially try all combinations until we locate the smallest one containing only identical characters.

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

First, consider all the smallest possible pieces of the string, each containing just a single character.
Next, examine all possible pieces containing two characters right next to each other.
Then, look at all pieces containing three characters right next to each other, and so on, systematically increasing the size of the piece we're inspecting.
For each piece we examine, check if all the characters within that piece are the same.
If we find a piece where all the characters are identical, we remember its size.
Continue checking every possible piece until we've examined all of them.
Finally, after looking at everything, we pick the smallest piece we found where all the characters were identical.

Code Implementation

def find_smallest_substring_identical_characters(text):
    shortest_length = float('inf')
    text_length = len(text)

    for start_index in range(text_length):
        for end_index in range(start_index, text_length):
            substring = text[start_index:end_index + 1]

            # Check if substring is empty
            if not substring:
                continue

            # First character to compare all other characters to
            first_character = substring[0]

            is_identical = True
            for character in substring:
                if character != first_character:
                    is_identical = False
                    break

            # Only update when characters are identical
            if is_identical:

                # Keep track of the current min length
                substring_length = len(substring)
                shortest_length = min(shortest_length, substring_length)

    # Handle empty or no substring of identical characters
    if shortest_length == float('inf'):
        return 0
    return shortest_length

Big(O) Analysis

Time Complexity

O(n³) – The brute force approach iterates through all possible substring lengths from 1 to n. For each length, it iterates through all possible starting positions in the string. Therefore, the outer loop iterates up to n times. The inner loop (for each substring length) also iterates up to n times. Inside the inner loop, we have to check if the characters in the substring are identical, which takes O(n) time in the worst case. Thus, the overall time complexity is O(n * n * n) which simplifies to O(n³).

Space Complexity

O(1) – The provided brute force approach checks substrings of various lengths, but it doesn't explicitly mention creating any auxiliary data structures that scale with the input size N (the length of the string). It only seems to store the size of the smallest substring found so far and perhaps a few variables to track the start and end of the current substring being examined. These variables take up constant space regardless of N, and no dynamic data structures are mentioned for storing intermediate substring results. Thus the auxiliary space complexity is constant.

Optimal Solution

Approach

To find the smallest substring with identical characters, we avoid checking every possible substring. Instead, we efficiently track where each character appears in the given text and use this knowledge to quickly identify potential substrings that meet the criteria.

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

First, record the positions of each unique character throughout the entire text.
Then, for each unique character, find the first and last occurrences.
The substring between the first and last occurrence of a character might be a candidate. So, calculate the length of this potential substring.
Check if this substring only contains the repeating character we are examining. If it does, it's a valid solution.
Keep track of the shortest valid substring you find. If a shorter valid substring is found, replace the current shortest substring with the new one.
After checking all the unique characters, the shortest substring found is the answer.

Code Implementation

def find_smallest_substring(input_string):
    left_pointer = 0
    right_pointer = 0
    smallest_length = float('inf')

    while right_pointer < len(input_string):
        # Expand the window
        current_window = input_string[left_pointer:right_pointer+1]
        character_set = set(current_window)

        # Check if the window contains only one type of character
        if len(character_set) == 1:
            substring_length = right_pointer - left_pointer + 1
            smallest_length = min(smallest_length, substring_length)
        else:
            # Shrink the window until only one character remains
            while len(set(input_string[left_pointer:right_pointer+1])) > 1 and left_pointer < right_pointer:
                left_pointer += 1

            # Re-evaluate window after shrinking
            if len(set(input_string[left_pointer:right_pointer+1])) == 1:
                substring_length = right_pointer - left_pointer + 1
                smallest_length = min(smallest_length, substring_length)

        right_pointer += 1

    if smallest_length == float('inf'):
        return 0
    else:
        return smallest_length

Big(O) Analysis

Time Complexity

O(n) – The algorithm first iterates through the input string of length n to record the positions of each unique character. This takes O(n) time. Then, for each unique character, it finds the first and last occurrences, which is O(1) because it can be done with a hashmap or similar data structure. The check if the substring between the first and last occurrences contains only the repeating character takes at most O(n) time in the worst case where the first and last character are at the extremeties. Since this substring check is only done for each unique character, the dominant cost becomes O(n) for the initial scan of the string, and the substring check is done less than n times. Therefore, the overall time complexity is O(n).

Space Complexity

O(N) – The dominant space usage comes from storing the positions of each unique character in the text. In the worst-case scenario, where all characters in the input string of length N are unique, we would store N positions. Therefore, the space required to store these positions scales linearly with the input size N. The other variables used, such as first and last occurrence indices and the shortest substring, take constant space.

Edge Cases

Case	How to Handle
Empty or null string	Return -1 immediately as there can be no substring with two occurrences of any character.
String with only one character	Return -1 since a single character can't have two occurrences.
String with two identical characters	Return 2 as it's the smallest possible substring.
String with all distinct characters	Iterate through all substrings to check for repeats; if none found, return -1 after checking all possibilities.
Very long string (scalability)	Use a sliding window or hash map approach to achieve O(n) time complexity instead of naive O(n^2) or O(n^3) substring comparison.
String with a single character repeated many times	The sliding window or hashmap should quickly identify the minimal substring (length 2 in this case).
String where a character repeats very late in the string	The algorithm must iterate until the end of the string if necessary to find that repeating character.
String with multiple characters repeating but only one creating the smallest substring	The solution needs to keep track of the minimum length found so far.