3456. Find Special Substring of Length K

Description

You are given a string s and an integer k.

Determine if there exists a substring of length exactly k in s that satisfies the following conditions:

The substring consists of only one distinct character (e.g., "aaa" or "bbb").
If there is a character immediately before the substring, it must be different from the character in the substring.
If there is a character immediately after the substring, it must also be different from the character in the substring.

Return true if such a substring exists. Otherwise, return false.

Example 1:

Input: s = "aaabaaa", k = 3

Output: true

Explanation:

The substring s[4..6] == "aaa" satisfies the conditions.

It has a length of 3.
All characters are the same.
The character before "aaa" is 'b', which is different from 'a'.
There is no character after "aaa".

Example 2:

Input: s = "abc", k = 2

Output: false

Explanation:

There is no substring of length 2 that consists of one distinct character and satisfies the conditions.

Constraints:

1 <= k <= s.length <= 100
s consists of lowercase English letters only.

Solutions

Solution 1: Two Pointers

The problem essentially asks us to find each segment of consecutive identical characters and then determine if there exists a substring of length \(k\). If such a substring exists, return \(\textit{true}\); otherwise, return \(\textit{false}\).

We can use two pointers \(l\) and \(r\) to traverse the string \(s\). When \(s[l] = s[r]\), move \(r\) to the right until \(s[r] \neq s[l]\). At this point, check if \(r - l\) equals \(k\). If it does, return \(\textit{true}\); otherwise, move \(l\) to \(r\) and continue traversing.

The time complexity is \(O(n)\), where \(n\) is the length of the string \(s\). The space complexity is \(O(1)\).

Python3JavaC++GoTypeScript

class Solution:
    def hasSpecialSubstring(self, s: str, k: int) -> bool:
        l, n = 0, len(s)
        while l < n:
            r = l
            while r < n and s[r] == s[l]:
                r += 1
            if r - l == k:
                return True
            l = r
        return False

class Solution {
    public boolean hasSpecialSubstring(String s, int k) {
        int n = s.length();
        for (int l = 0, cnt = 0; l < n;) {
            int r = l + 1;
            while (r < n && s.charAt(r) == s.charAt(l)) {
                ++r;
            }
            if (r - l == k) {
                return true;
            }
            l = r;
        }
        return false;
    }
}

class Solution {
public:
    bool hasSpecialSubstring(string s, int k) {
        int n = s.length();
        for (int l = 0, cnt = 0; l < n;) {
            int r = l + 1;
            while (r < n && s[r] == s[l]) {
                ++r;
            }
            if (r - l == k) {
                return true;
            }
            l = r;
        }
        return false;
    }
};

func hasSpecialSubstring(s string, k int) bool {
    n := len(s)
    for l := 0; l < n; {
        r := l + 1
        for r < n && s[r] == s[l] {
            r++
        }
        if r-l == k {
            return true
        }
        l = r
    }
    return false
}

function hasSpecialSubstring(s: string, k: number): boolean {
    const n = s.length;
    for (let l = 0; l < n; ) {
        let r = l + 1;
        while (r < n && s[r] === s[l]) {
            r++;
        }
        if (r - l === k) {
            return true;
        }
        l = r;
    }
    return false;
}

3456. Find Special Substring of Length K

Description

Solutions

Solution 1: Two Pointers

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