LC 3456: Posting cuz it's super intuitive.

Find the problem description here. If you would like to read the solution in leetcode platform, find it here.

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Intuition

If you think about it, it is just calculating the frequency of same consecutive characters. The frequency is all '1' if no characters are same consecutively.

Example:
index: 01234567
string: baaabaaa
k = 3
freq = 1

we start from index 1,
freq resets to 1 when i = 1 as a != b.

before resetting, check if freq == k
freq = 2 when i = 2 as a == a.
freq = 3 when i = 3 as a == a.
freq resets to 1 when i = 4 as b != a.
before resetting, check if freq == k, it is. so, return true.

done! no isolation checks or anything.

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Approach

• Start with the current count '1', as a single character's frequency is '1'.
• If the next characters is same as before, increment the count.
• If it is not equal, simple, check whether the frequency is 'k', if it is.. then you found your special string. Yes, this works.
• If it is not 'k', then reset the current count to '1' again.

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Complexity

• Time complexity: $O(n)$

• Space complexity: $O(1)$

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Code

class Solution {
public:
    bool hasSpecialSubstring(string s, int k) {
        int n = s.size();
        int currCount = 1;
        for (int i = 1; i < n; ++i) {
            // calculating the frequency of consecutive characters
            if (s[i] == s[i - 1])   currCount++;
            else {
                // if they reached k then, special
                if (currCount == k) return true;
                // reset
                currCount = 1;
            }
        }
        // checking any trailing consecutive characters
        if (currCount == k) return true;

        return false;
    }
};

good luck!