How do you solve Substring With Largest Variance on LeetCode?

Substring With Largest Variance (LeetCode #2272) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(1) space complexity. Key insight: The variance is maximized by focusing on pairs of characters.

What is the brute force solution for Substring With Largest Variance?

The brute force approach for Substring With Largest Variance is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible substrings of the input string and calculating the variance for each substring. This method is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Substring With Largest Variance?

Substring With Largest Variance uses the following concepts: Hash Table, String, Dynamic Programming, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Substring With Largest Variance?

Think about character pairs and how they interact. Practice counting techniques in strings. Be ready to explain your thought process clearly.

What are common mistakes when solving Substring With Largest Variance?

Not considering all pairs of distinct characters. Failing to reset counts when the variance condition is violated.

#2272

Substring With Largest Variance

Hard

Hash TableStringDynamic ProgrammingEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(1)

The optimal solution focuses on pairs of characters and uses a modified counting approach to efficiently calculate the variance. By treating one character as +1 and another as -1, we can track the maximum variance using a single pass through the string.

⚙️

Algorithm

3 steps

1Step 1: Iterate through all pairs of distinct characters in the string.
2Step 2: For each pair, traverse the string, treating one character as +1 and the other as -1.
3Step 3: Maintain a running total and track the maximum variance encountered, resetting when the total becomes negative.

solution.py21 lines

1# Full working Python code
2
3def largestVariance(s):
4    max_variance = 0
5    unique_chars = set(s)
6    for a in unique_chars:
7        for b in unique_chars:
8            if a != b:
9                count_a = count_b = 0
10                for char in s:
11                    if char == a:
12                        count_a += 1
13                    elif char == b:
14                        count_b += 1
15                    if count_b > count_a:
16                        count_a = count_b = 0
17                    max_variance = max(max_variance, count_a - count_b)
18    return max_variance
19
20# Example usage
21print(largestVariance('aababbb'))  # Output: 3

ℹ

Complexity note: The time complexity is O(n²) because we iterate through all pairs of distinct characters, and for each pair, we traverse the string once. The space complexity is O(1) since we only use a fixed amount of extra space regardless of input size.

1The variance is maximized by focusing on pairs of characters.
2Resetting counts when one character exceeds another helps track the variance.

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