What is the brute force solution for Count Substrings Starting and Ending with Given Character?

The brute force approach for Count Substrings Starting and Ending with Given Character is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible substrings of the given string and counting those that start and end with the specified character. 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 Count Substrings Starting and Ending with Given Character?

Count Substrings Starting and Ending with Given Character uses the following concepts: Math, String, Counting. The recommended patterns to study are: Counting, Combinatorial Mathematics.

What are the interview tips for Count Substrings Starting and Ending with Given Character?

Always think about the brute-force solution first to understand the problem deeply. Look for patterns in the problem that can lead to a more efficient solution. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Count Substrings Starting and Ending with Given Character?

Forgetting to include single-character substrings. Not considering edge cases, like when the character does not appear in the string.

#3084

Count Substrings Starting and Ending with Given Character

Medium

MathStringCountingCountingCombinatorial Mathematics

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)

Instead of checking all possible substrings, we can count the occurrences of the character and use combinatorial mathematics to calculate the number of valid substrings directly. This is much more efficient.

⚙️

Algorithm

2 steps

1Step 1: Count the total occurrences of the character c in the string s. Let this count be m.
2Step 2: Use the formula m * (m + 1) / 2 to calculate the total number of substrings that start and end with c. This works because each pair of occurrences can form a substring.

solution.py8 lines

1# Full working Python code
2
3def count_substrings(s, c):
4    m = s.count(c)
5    return m * (m + 1) // 2
6
7# Example usage
8print(count_substrings('abada', 'a'))  # Output: 6

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once to count occurrences of c. The space complexity is O(1) since we use a constant amount of extra space.

1Counting occurrences can simplify problems involving substrings.
2Using combinatorial mathematics can significantly reduce time complexity.

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