How do you solve Count Number of Homogenous Substrings on LeetCode?

Count Number of Homogenous Substrings (LeetCode #1759) 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: Homogenous substrings can be counted based on the length of contiguous segments of the same character.

What is the time complexity of Count Number of Homogenous Substrings?

The optimal solution for Count Number of Homogenous Substrings runs in O(n) time and uses O(1) space. This complexity is linear because we are making a single pass through the string, counting segments without generating substrings.

What is the brute force solution for Count Number of Homogenous Substrings?

The brute force approach for Count Number of Homogenous Substrings 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 checking if they are homogenous. 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 Count Number of Homogenous Substrings?

Count Number of Homogenous Substrings uses the following concepts: Math, String. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Count Number of Homogenous Substrings?

Always think about the constraints and how they affect your approach. Discuss your thought process out loud to help the interviewer understand your reasoning. Consider edge cases, such as strings with all identical characters or all unique characters.

What are common mistakes when solving Count Number of Homogenous Substrings?

Failing to consider the modulo operation when counting large numbers. Not resetting the current length when encountering a different character.

#1759

Count Number of Homogenous Substrings

Medium

MathStringSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal approach leverages the fact that consecutive characters contribute to homogenous substrings in a predictable way. Instead of generating all substrings, we can count them directly based on the length of contiguous characters.

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Algorithm

5 steps

1Step 1: Initialize a counter for the total number of homogenous substrings and a variable to track the length of the current homogenous segment.
2Step 2: Iterate through the string, comparing each character with the previous one.
3Step 3: If they are the same, increment the length of the current segment. If not, reset the length to 1.
4Step 4: For each segment of length k, add k * (k + 1) / 2 to the total count.
5Step 5: Return the total count modulo 10^9 + 7.

solution.py13 lines

1# Full working Python code
2
3def countHomogenous(s):
4    count = 0
5    current_length = 1
6    mod = 10**9 + 7
7    for i in range(1, len(s)):
8        if s[i] == s[i - 1]:
9            current_length += 1
10        else:
11            current_length = 1
12        count = (count + current_length) % mod
13    return count

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Complexity note: This complexity is linear because we are making a single pass through the string, counting segments without generating substrings.

1Homogenous substrings can be counted based on the length of contiguous segments of the same character.
2The formula for counting substrings from a segment of length k is k * (k + 1) / 2.

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