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Smallest Substring With Identical Characters I

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StringsSliding Windows

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:

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:

s = "0000", numOps = 2

Output: 1

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

Example 3:

s = "0101", numOps = 0

Output: 1

Constraints:

  • 1 <= n == s.length <= 1000
  • s consists only of '0' and '1'.
  • 0 <= numOps <= n

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:

  1. What is the expected data type of the input string, and can it contain characters other than lowercase English letters?
  2. Is the input string guaranteed to contain at least one character?
  3. If there are multiple substrings with identical characters of the same smallest length, should I return any one of them?
  4. If the input string only contains a single character, should I return the original string itself?
  5. What should be returned if the input string is empty or null?

Brute Force Solution

Approach

The brute force approach to finding the smallest substring with identical characters involves examining every possible substring. We achieve this by systematically checking each potential substring and verifying if it meets the problem's conditions. This exhaustive search guarantees we'll find the shortest valid substring.

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

  1. Start by looking at all the smallest possible pieces of the text, which are single characters.
  2. Check if any of these single characters satisfy the problem condition (i.e., if the single character is identical to itself).
  3. Then, look at all possible pairs of adjacent characters.
  4. Check if each pair meets the problem condition (i.e., if both characters in the pair are the same).
  5. Next, look at all possible pieces of three adjacent characters, then four, and so on, gradually increasing the size of the pieces you are checking.
  6. For each possible piece, check if it satisfies the problem condition (i.e., if all characters in that piece are identical).
  7. Keep track of the size of the smallest piece that you found that satisfies the condition.
  8. After checking all possible pieces of all possible sizes, the smallest size you kept track of is the answer.

Code Implementation

def find_smallest_substring_brute_force(input_string):
    string_length = len(input_string)
    smallest_substring_length = float('inf')

    # Iterate through all possible substring lengths
    for substring_length in range(1, string_length + 1):
        # Iterate through all possible starting positions for each substring length
        for start_index in range(string_length - substring_length + 1):
            substring = input_string[start_index:start_index + substring_length]

            # Check if all characters in the substring are identical
            if len(set(substring)) == 1:

                #Update the smallest substring length if a shorter one is found
                smallest_substring_length = substring_length
                break

        if smallest_substring_length != float('inf'):
            break

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

Big(O) Analysis

Time Complexity
O(n^3)The brute force approach iterates through all possible substring lengths from 1 to n, where n is the length of the input string. For each length, it iterates through all possible starting positions within the string. For each substring, it checks if all characters within that substring are identical, requiring a linear scan of the substring. Thus, the time complexity is O(n) for checking each substring, multiplied by the number of possible substrings which is O(n^2). This means the total operations approximate to O(n * n^2), which simplifies to O(n^3).
Space Complexity
O(1)The described brute force approach doesn't utilize any auxiliary data structures. It only involves iterating through substrings and comparing characters directly within the input string. No extra space is used to store intermediate substrings or track visited states. Therefore, the space complexity remains constant irrespective of the input string's length N.

Optimal Solution

Approach

To find the shortest segment with all identical characters, we don't need to check every possible segment. Instead, we can cleverly jump from one possible solution to another, quickly narrowing down the possibilities until we find the absolute shortest one.

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

  1. Look at the input. If it has less than 3 characters, the answer is obvious.
  2. Start by checking if any three consecutive characters are the same. If so, that's a segment of length 3, and a possible solution.
  3. Go through the input, looking for any place where two characters are the same. Remember the location of the first one.
  4. Once we find a pair, keep looking ahead to see how many more characters are the same after the initial pair. Basically, find the full 'run' of identical characters.
  5. If the 'run' of same characters is shorter than a previous shortest segment that you may have seen, ignore it.
  6. If it is shorter than any previous shortest segment, then remember the run of same characters and its length as the current shortest.
  7. Continue scanning the string and repeating steps 3-6 until we reach the end.
  8. Return the length of the shortest 'run' you found.

Code Implementation

def find_shortest_segment(input_string):
    string_length = len(input_string)

    if string_length < 3:
        return string_length

    shortest_length = float('inf')

    # Check for initial consecutive identical characters.
    for i in range(string_length - 2):
        if input_string[i] == input_string[i+1] == input_string[i+2]:
            shortest_length = 3
            break

    for i in range(string_length - 1):
        if input_string[i] == input_string[i+1]:

            # Found a pair, find the entire run length.
            current_length = 2
            start_index = i

            while i + current_length < string_length and input_string[start_index] == input_string[i + current_length]:
                current_length += 1

            # Update shortest length if needed.
            if current_length < shortest_length:
                shortest_length = current_length

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

Big(O) Analysis

Time Complexity
O(n)The algorithm iterates through the input string of length n to find identical characters. The initial check for consecutive identical characters has a cost proportional to n. The algorithm then iterates through the string, looking for pairs, and extends the 'run' of identical characters. While there's a nested 'while' loop implicitly, each character is only processed a constant number of times. Therefore, the overall time complexity is dominated by the single pass through the string, resulting in O(n).
Space Complexity
O(1)The algorithm uses a constant number of variables to store the location of the first same character, the length of the current run of identical characters, and the length of the shortest run found so far. No auxiliary data structures like arrays, hash maps, or lists are created that scale with the input size N (the length of the input string). Therefore, the space complexity remains constant regardless of the input string's length.

Edge Cases

CaseHow to Handle
Null or empty input stringReturn 0 (or appropriate sentinel value indicating no substring found) immediately, preventing null pointer exceptions or invalid operations.
Input string with only one characterReturn 1, as a single character is a substring with identical characters.
Input string with all identical charactersReturn 1, as any single character substring is the smallest.
Input string with length approaching maximum allowed string lengthEnsure the algorithm has a time complexity that can handle large strings (e.g., O(n) or O(n log n)).
String contains Unicode or other non-ASCII charactersEnsure the character comparison and storage mechanisms (e.g., hash maps) correctly handle Unicode characters.
No substring with identical characters exists (e.g., 'abc')Return 0, indicating that no valid substring exists.
String contains only whitespace charactersTreat whitespace as any other character and find smallest substring of identical whitespace characters, returning 1 if it exists, 0 otherwise.
Integer overflow in length calculations (if applicable)Use appropriate data types (e.g., long instead of int) to prevent integer overflows when calculating substring lengths.