[Cracking The Coding Interview] 1.4: Palindrome Permutation

[jtkaufman737]

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.)

[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);

[gauravchl]

https://repl.it/@gauravchl/Palindrome-Permutation

function isPalindrome(str) {
  const charCounts = {};
  const inputString = str.replace(/ /g, '');

  for(let i = 0; i < inputString.length; i++) {
    const char = inputString[i].toLowerCase();
    charCounts[char] = (charCounts[char] || 0) + 1;
  }

  const oddsCount = Object
    .values(charCounts)
    .reduce((a, b) => b % 2 === 1 ? a + 1 : a, 0)

  return oddsCount <= 2
}

isPalindrome('asd a sd') // true
isPalindrome('level') // true
isPalindrome('asdfgr') // false
isPalindrome('Tact Coa') // true