How do you solve Replace the Substring for Balanced String on LeetCode?

Replace the Substring for Balanced String (LeetCode #1234) 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: A string is balanced when each character appears exactly n/4 times.

What is the brute force solution for Replace the Substring for Balanced String?

The brute force approach for Replace the Substring for Balanced String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible substrings of the input string and calculates the minimum length of a substring that can be replaced to achieve a balanced string. This 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 Replace the Substring for Balanced String?

Replace the Substring for Balanced String uses the following concepts: String, Sliding Window. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Replace the Substring for Balanced String?

Always consider edge cases, such as when the string is already balanced. Explain your thought process clearly while coding; it helps interviewers understand your approach. Practice sliding window problems to get comfortable with the technique.

What are common mistakes when solving Replace the Substring for Balanced String?

Not checking if the string is already balanced before starting the replacement process. Using nested loops without considering the efficiency of the solution.

#1234

Replace the Substring for Balanced String

Medium

StringSliding WindowSliding 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 substring that can be replaced. This method allows us to maintain a window of characters and adjust it dynamically, ensuring we only check necessary parts of the string.

⚙️

Algorithm

6 steps

1Step 1: Count the occurrences of each character in the string.
2Step 2: Calculate the target count for each character (n / 4).
3Step 3: Use two pointers to create a sliding window over the string.
4Step 4: Expand the right pointer to include characters until the window is valid (all characters are within the target count).
5Step 5: Once valid, move the left pointer to minimize the window size while maintaining validity.
6Step 6: Track the minimum length of the valid window.

solution.py16 lines

1def balancedString(s):
2    from collections import Counter
3    n = len(s)
4    target = n // 4
5    count = Counter(s)
6    if all(count[c] == target for c in 'QWER'):
7        return 0
8    left = 0
9    min_length = n
10    for right in range(n):
11        count[s[right]] -= 1
12        while all(count[c] <= target for c in 'QWER'):
13            min_length = min(min_length, right - left + 1)
14            count[s[left]] += 1
15            left += 1
16    return min_length

ℹ

Complexity note: The time complexity is O(n) because we traverse the string with two pointers, ensuring each character is processed a limited number of times. The space complexity is O(1) as we only use a fixed-size array for counting characters.

1A string is balanced when each character appears exactly n/4 times.
2Using a sliding window allows us to efficiently find the minimum substring to replace.

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