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4030. Check ASCII Palindromic

Description

You are given a string s consisting of lowercase English letters.

Construct a binary string by replacing each character in s with the 8-bit binary representation of its ASCII value, including leading zeros, while preserving the original order of the characters.

Return true if the resulting binary string is a palindrome. Otherwise, return false.

 

Example 1:

Input: s = "ff"

Output: true

Explanation:

  • The ASCII value of f is 102, whose 8-bit binary representation is 01100110.
  • Thus, the binary string is 0110011001100110.
  • Since this binary string is a palindrome, the output is true.

Example 2:

Input: s = "leet"

Output: false

Explanation:

  • The ASCII values of l, e, e, and t are 108, 101, 101, and 116, respectively.
  • Their 8-bit binary representations are 01101100, 01100101, 01100101, and 01110100.
  • Thus, the binary string is 01101100011001010110010101110100.
  • Since this binary string is not a palindrome, the output is false.

 

Constraints:

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

Solutions

Solution 1: Simulation

Thinking

Each character expands to a fixed \(8\)-bit string, including leading zeros, and we test whether the concatenation is a palindrome. \(n\) is small enough that we need not rewrite the test in terms of character pairs.

Build \(t\) in order and compare it with its reverse.

Following the problem statement, we replace each character of \(s\) with the \(8\)-bit binary representation of its ASCII value (including leading zeros), concatenate them in order to obtain a binary string \(t\), and then check whether \(t\) is a palindrome.

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

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class Solution:
    def isPalindromic(self, s: str) -> bool:
        t = ''.join(format(ord(c), '08b') for c in s)
        return t == t[::-1]

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class Solution {
    public boolean isPalindromic(String s) {
        StringBuilder t = new StringBuilder();
        for (char c : s.toCharArray()) {
            String b = Integer.toBinaryString(c);
            t.append("0".repeat(8 - b.length())).append(b);
        }
        return t.toString().equals(t.reverse().toString());
    }
}

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class Solution {
public:
    bool isPalindromic(string s) {
        string t;
        for (unsigned char c : s) {
            for (int i = 7; i >= 0; --i) {
                t += char('0' + ((c >> i) & 1));
            }
        }
        return ranges::equal(t, t | views::reverse);
    }
};

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func isPalindromic(s string) bool {
    var t []byte
    for _, c := range []byte(s) {
        for i := 7; i >= 0; i-- {
            t = append(t, '0'+((c>>i)&1))
        }
    }
    for i := range t[:len(t)/2] {
        if t[i] != t[len(t)-1-i] {
            return false
        }
    }
    return true
}

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function isPalindromic(s: string): boolean {
    const t = [...s].map(c => c.charCodeAt(0).toString(2).padStart(8, '0')).join('');
    return t === [...t].reverse().join('');
}

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