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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:

1. The substring consists of only one distinct character (e.g., "aaa" or "bbb").
2. If there is a character immediately before the substring, it must be different from the character in the substring.
3. 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 $\mathit{\text{true}}$; otherwise, return $\mathit{\text{false}}$.

We can use two pointers $l$ and $r$ to traverse the string $s$. When $s[l]=s[r]$, move $r$ to the right until $s[r]\ne s[l]$. At this point, check if $r-l$ equals $k$. If it does, return $\mathit{\text{true}}$; 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 $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  };
  
  

• 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
  
  

• 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;
  }
  
  

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