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Barsik Samogonov

Barsik Samogonov

· a month ago

Constraint mismatch: 1 <= s.length stated, but test case passes "" (length 0)

The problem states:

1 <= s.length <= 2 * 10^5

However, one of the test cases passes an empty string s = "". I confirmed this by logging at the very start of the function:

if (s === "") {

console.log(logging test case input "${s}" and length ${s.length});

}

Output: logging test case input "" and length 0

Should the constraints be updated to 0 <= s.length, or is this test case an error in the suite?

Show 1 reply
Sushant Gautam

Sushant Gautam

· 2 months ago

class Solution:
  def isPalindrome(self, s: str) -> bool:
    cleaned_l = [char.lower() for char in s if char.isalpha() or char.isdigit()]
    # two pointer
    l = 0
    r = len(cleaned_l) - 1
    while l < r:
      if cleaned_l[l] != cleaned_l[r]:
        return False
      l += 1
      r -= 1

    return True
Show 1 reply
Ana Calin

Ana Calin

· a year ago

Here's my solution for the palindrome problem that compiles and runs successfully on my editor:

package main

import (
  "fmt"
  "regexp"
  "strings"
)

type Solution struct {
}

func (sol *Solution) isPalindrome(s string) bool {
  first, last := 0, len(s)-1
  for first < last {
    if s[first] != s[last] {
      return false
    }
    first++
    last--
  }
  return true
}

func normalise(s string) string {
  toLowerStr := strings.ToLower(s)
  reg := regexp.MustCompile(`[^a-z]`)
  transformedStr := reg.ReplaceAllString(toLowerStr, "")
  return transformedStr
}

func main() {
  str := "A man, a plan, a canal, Panama!"
  normalisedStr := normalise(str)
  solution := Solution{}
  fmt.Println("Is my string a palindrome?", solution.isPalindrome(normalisedStr))
}

However

Show 1 reply
Amber Wolf

Amber Wolf

· a year ago

    public bool IsPalindrome(string s)
    {
        // use Regex to easily remove special characters and spaces from our string
        var trimmedString = Regex.Replace(s, "[^a-zA-Z0-9]", "").ToLower();
        // create an array from the string so we can iterate through it
        trimmedString.ToCharArray();
        // we need to track both the Forward facing and Backward facing iterations of this array so I am initializing the index starting at the end of the array here.
        int j = trimmedString.Length - 1;
        // each time we loop through our array we will increase i and decrease j to compare both the forward and backward letter of the string.
        for (int i = 0; i < trimmedString.Length; i++, j--)
        {
            if (trimmedString[i] != trimmedString[j])
Show 1 reply
Akshay Hedau

Akshay Hedau

· 3 years ago

class Solution {
  public boolean isPalindrome(String s) {
    // TODO: Write your code here
    s = s.toLowerCase();
    int st = 0;
    s = s.toLowerCase();
    int ed = s.length() - 1;
    char[] curS = s.toCharArray();

    while (st < ed) {
      char curSt = curS[st];
      char curEd = curS[ed];

      if (curSt > 'a' && curSt < 'z' && curEd > 'a' && curEd < 'z') {
        if (curSt != curEd) return false;
        st++;
        ed--;
      } else if (curSt > 'a' && curSt < 'z') {
        ed--;
      } else if (curEd > 'a' && curEd < 'z') {
        st++;
      } else {
        st++;
        ed--;
      }
    }
    return true; 
  }
}
Show 1 reply
T

Thai Minh

· 3 years ago

class Solution:
  def isPalindrome(self, s: str) -> bool:
    # TODO: Write your code here
    str_array = []
    for char in s:
      if char.isalpha() or char.isdigit():
        str_array.append(char.lower())
    front, back = 0, len(str_array) - 1
    while front < back:
      if str_array[front] != str_array[back]:
        return False
      front += 1
      back -= 1
    return True
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