Welcome to Subscribe On Youtube

3083. Existence of a Substring in a String and Its Reverse

Description

Given a string s, find any substring of length 2 which is also present in the reverse of s.

Return true if such a substring exists, and false otherwise.

 

Example 1:

Input: s = "leetcode"

Output: true

Explanation: Substring "ee" is of length 2 which is also present in reverse(s) == "edocteel".

Example 2:

Input: s = "abcba"

Output: true

Explanation: All of the substrings of length 2 "ab", "bc", "cb", "ba" are also present in reverse(s) == "abcba".

Example 3:

Input: s = "abcd"

Output: false

Explanation: There is no substring of length 2 in s, which is also present in the reverse of s.

 

Constraints:

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

Solutions

Solution 1: Hash Table or Array

We can use a hash table or a two-dimensional array $st$ to store all substrings of length $2$ of the reversed string $s$.

Then we traverse the string $s$. For each substring of length $2$, we check whether it has appeared in $st$. If it has, we return true. Otherwise, we return false after the traversal.

The time complexity is $O(n)$ and the space complexity is $O(|\Sigma|^2)$. Here, $n$ is the length of the string $s$, and $\Sigma$ is the character set of the string $s$. In this problem, $\Sigma$ consists of lowercase English letters, so $|\Sigma| = 26$.

  • class Solution {
    public boolean isSubstringPresent(String s) {
        boolean[][] st = new boolean[26][26];
        int n = s.length();
        for (int i = 0; i < n - 1; ++i) {
            st[s.charAt(i + 1) - 'a'][s.charAt(i) - 'a'] = true;
        }
        for (int i = 0; i < n - 1; ++i) {
            if (st[s.charAt(i) - 'a'][s.charAt(i + 1) - 'a']) {
                return true;
            }
        }
        return false;
    }
}

  • class Solution {
public:
    bool isSubstringPresent(string s) {
        bool st[26][26]{};
        int n = s.size();
        for (int i = 0; i < n - 1; ++i) {
            st[s[i + 1] - 'a'][s[i] - 'a'] = true;
        }
        for (int i = 0; i < n - 1; ++i) {
            if (st[s[i] - 'a'][s[i + 1] - 'a']) {
                return true;
            }
        }
        return false;
    }
};

  • class Solution:
    def isSubstringPresent(self, s: str) -> bool:
        st = {(a, b) for a, b in pairwise(s[::-1])}
        return any((a, b) in st for a, b in pairwise(s))


  • func isSubstringPresent(s string) bool {
	st := [26][26]bool{}
	for i := 0; i < len(s)-1; i++ {
		st[s[i+1]-'a'][s[i]-'a'] = true
	}
	for i := 0; i < len(s)-1; i++ {
		if st[s[i]-'a'][s[i+1]-'a'] {
			return true
		}
	}
	return false
}

  • function isSubstringPresent(s: string): boolean {
    const st: boolean[][] = Array.from({ length: 26 }, () => Array(26).fill(false));
    for (let i = 0; i < s.length - 1; ++i) {
        st[s.charCodeAt(i + 1) - 97][s.charCodeAt(i) - 97] = true;
    }
    for (let i = 0; i < s.length - 1; ++i) {
        if (st[s.charCodeAt(i) - 97][s.charCodeAt(i + 1) - 97]) {
            return true;
        }
    }
    return false;
}

All Problems

All Solutions