1.4 Palindrome Permutation

[mcsmithers]

Palindrome Permutation: Given a string, write a function to check if it is a permutation of a palindrome. A palindrome is a word or phrase that is the same forwards and backwards. A permutation is a rearrangement of letters. The palindrome does not need to be limited to just dictionary words.

EXAMPLE
Input: Tact Coa
Output: True (permutations: "taco cat", "atco eta", etc.)

[megantaylor]

https://repl.it/@megantaylor/PalindromePermutation

function isPalindromicPermutation(str) {
  let chars = {};
  let lowerCaseStr = str.toLowerCase();
  if(str.length < 1) {
    return false;
  }
  for(let i = 0; i < str.length; i++){
    if(lowerCaseStr[i] !== " ") {
      if(chars[lowerCaseStr[i]]) {
      chars[lowerCaseStr[i]] += 1;
    } else {
      chars[lowerCaseStr[i]] = 1;
    }
    } 
  }
  let numOddChars = Object.values(chars).filter(char => char % 2 !== 0);
  return numOddChars.length > 1 ? false : true; 
}


function isPalindromicPermutationFancy(string) {
let oddCharacters = [...new Set(string)].filter( (char)=> {
  let numberOfOccurences = [...string].filter( (letter) => {
    return letter.toLowerCase() === char.toLowerCase() && letter !== " ";
  });
  return numberOfOccurences.length % 2 === 1;
});

if( oddCharacters.length === 1 || oddCharacters.length === 0 ){
  return true;
} else {
  return false;
}
}

[jtkaufman737]

https://repl.it/@jtkaufman737/QuickwittedMundaneWatchdog

let palindrome = "taco cat"; // should pass 
let palindrome2 = "aa"; // should pass 
let palindrome3 = "ra ce car"; // should pass 
let palindrome4 = "si de car"; // should fail 


function isPalindrome(str) {
  let keys = {}, vals = [];

  Array.from(str.split(" ").join("")).map(m => { 
    keys[m] = 0;  // initial object with letters as keys, each count = 0 
  })

  Array.from(str).map(m => { 
    keys[m] += 1; // For all array items, increments the count of each letter
  })

  let odds = Object.values(keys).filter(f => {
    return f % 2 === 1; // if MORE THAN ONE odd number is there, it isn't a palindrome
  })

  odds.length > 1 ? console.log("Sorry, not a palindrome") : console.log("Congrats, its a palindrome!");

}

isPalindrome(palindrome);
isPalindrome(palindrome2);
isPalindrome(palindrome3);
isPalindrome(palindrome4);

[jtkaufman737]

ooh looking at @megantaylor mine would def fail on capital letters! Good thinking

[jtkaufman737]

here's some python: https://repl.it/@jtkaufman737/FrenchZealousTechnologies

def testPalin(string):
  string = list(string.lower()) # eliminate case issues 
  odds = 0
  # print(string)
  
  for str in string:
    if str not in ' ':
      keys = { str : 0 }
      # print(keys)

  for str in string:
    if str not in ' ':
      keys[str] = keys.get(str, 0) + 1
      # print(keys)

  for val in keys.values():
    if val % 2 == 1:
      odds += 1 

  if odds > 1:
    print("Sorry, not a palindrome")
  else:
    print("Yay, you've got a palindrome!")


testPalin("tacoc AT") # should pass 
testPalin("AaBbCcDd") # should pass
testPalin("AaAbBbCcC") # should fail  
testPalin("racecar") # should pass 
testPalin("sidecar") # should fail 

[rmorabia]

https://repl.it/@rmorabia/ThoughtfulBoringWorkspace

O(n^2)

function isPalindrome(str) {
  let string = str
    .replace(/[^a-z]/gi, "")
    .toLowerCase()
    .split("");
  let justOne = 0;
  if (string.length <= 1) {
    return "idk i guess this is a palindrome leave me alone 2k18";
  }
  for (let i = 0; i < string.length; i++) {
    let count = 0;
    for (let z = 0; z < string.length; z++) {
      if (string[i] === string[z]) {
        count++;
      }
    }
    if (count % 2 !== 0 && count !== 1) {
      return "sry m8 u failed";
    } else if (count === 1) {
      justOne++;
      if (justOne >= 2) {
        return "sry m8 u failed";
      }
    }
  }
  return "w0w";
}

console.log(isPalindrome("Tact Coa"));
console.log(isPalindrome("abcd"));
console.log(isPalindrome("tttt"));
console.log(isPalindrome("taco bell Lleb oCat"));
console.log(isPalindrome("taco bell Llbb oCat"));

EDIT: THE SOLUTION BELOW IS TO CHECK IF A STRING IS A PALINDROME! Because apparently I forgot the problem halfway through!

https://repl.it/@rmorabia/EmptyUnrealisticProjector

function isPalindrome(str) {
  let string = str
    .replace(/[^a-z]/gi, "")
    .toLowerCase()
    .split("");
  if (string.length <= 1) {
    return "do a better string next time";
  }
  for (let i = 0; i < string.length;) {
    if (string[0] == string[string.length - 1]) {
      string.shift();
      string.pop();
    } else {
      return "do a better string next time"
    }
  }
  if (string.length <= 1) {
    return "w0w much p4l1ndr0m3";
  } else {
    return "do a better string next time";
  }
}

console.log(isPalindrome("madam"))
console.log(isPalindrome("taco bell Lleb oCat"));
console.log(isPalindrome("taco bell Llbb oCat"));

[lpatmo]

https://repl.it/@lpatmo/FilthyTimelySale

function palPerm(str) {
  //Check if there is an even number of each possible letter
  //It is a permutation of a palindrome IF after you remove each pair of letters, there is 0 or 1 chars remaining

  //Create temp string
  //Loop through original str
  //If I do not see it in temp, add it to temp. 
  //If I see it in temp, remove it from temp.
  //At the end, the temp string will hold ALL of the odd counts of chars.
  //If the length of temp string is greater than 1, return false.
  let temp = [];
  for (let i = 0; i < str.length; i++) {
    let charPosition = temp.indexOf(str[i]);
    if ( charPosition > -1 ) {
      temp.splice(charPosition, 1);
    } else if (str[i] !== ' ') {
      temp.push(str[i]);
    }
  }
  //console.log(temp)
  return temp.length < 2
}

console.log(palPerm("tcao cat"))

[mcsmithers]

Xtina's solution

const testStr = 'madam';
const testStr2 = 'lady';

const testStr = 'madam';
const testStr2 = 'lady';

function checkPalindromes(myStr) {

  // Reverse the string  
  const reverseStr = myStr.split('').reverse().join('');
  if (reverseStr === myStr) {
    console.log(myStr + ' is a palindrome');
  }
  else {
    console.log(myStr + ' is not a plaindrome');
  }
  return reverseStr;
}

checkPalindromes(testStr);