How do you solve Smallest Substring With Identical Characters I on LeetCode?

Smallest Substring With Identical Characters I (LeetCode #3398) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using a sliding window can significantly reduce the time complexity.

What is the brute force solution for Smallest Substring With Identical Characters I?

The brute force approach for Smallest Substring With Identical Characters I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible substring of the binary string and counting how many flips are needed to make all characters in that substring identical. This method is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Smallest Substring With Identical Characters I?

Smallest Substring With Identical Characters I uses the following concepts: Array, Binary Search, Enumeration. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Smallest Substring With Identical Characters I?

Always clarify the problem constraints before diving into coding. Think about edge cases and how they might affect your solution. Discuss your thought process with the interviewer to showcase your problem-solving skills.

What are common mistakes when solving Smallest Substring With Identical Characters I?

Not considering edge cases where numOps is 0. Failing to optimize the substring checks in the brute-force approach.

#3398

Smallest Substring With Identical Characters I

Hard

ArrayBinary SearchEnumerationSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a sliding window approach to efficiently find the minimum length of the longest substring with identical characters after performing the allowed number of flips. This method is faster because it avoids redundant checks by maintaining a window of valid substrings.

⚙️

Algorithm

4 steps

1Step 1: Initialize two pointers (left and right) to represent the current window of characters.
2Step 2: Expand the right pointer to include more characters until the number of flips exceeds numOps.
3Step 3: If flips exceed numOps, move the left pointer to reduce the window size until flips are within the limit.
4Step 4: Keep track of the maximum length of the window that can be made identical with the allowed flips.

solution.py22 lines

1# Full working Python code
2
3def min_length_after_flips(s, numOps):
4    left = 0
5    count_flips = 0
6    max_length = float('inf')
7
8    for right in range(len(s)):
9        if s[right] == '0':
10            count_flips += 1
11
12        while count_flips > numOps:
13            if s[left] == '0':
14                count_flips -= 1
15            left += 1
16
17        max_length = min(max_length, right - left + 1)
18
19    return max_length
20
21# Example usage
22print(min_length_after_flips('000001', 1))  # Output: 2

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once with the two pointers. The space complexity is O(1) since we are using a fixed amount of extra space.

1Using a sliding window can significantly reduce the time complexity.
2Understanding how to count flips efficiently is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.