const areAnagram = (str1, str2) => str1.toLowerCase().split('').sort().join('') === str2.toLowerCase().split('').sort().join('');
// Examples
areAnagram('listen', 'silent'); // true
areAnagram('they see', 'the eyes'); // true
areAnagram('node', 'deno'); // true
/*
An anagram can be obtained by rearranging
the letters of another word. For instance,
"anagram" is an anagram of "nagaram".
This implementation determines if one string
is anagram of another. The two strings are
assumed to be composed of lowercase alphabetic
letters only.
Time complexity: O(n)
Space complexity: O(1)
*/
// Return true if str1 is anagram of str2
// Or false otherwise
function isAnagram(str1, str2) {
// Two anagrams must have same length
if (str1.length !== str2.length) {
return false;
}
// Count nb of occurrences of characters
// in str1. Only 26 countes are needed
// since str1 has only lowercase letters
const counter = new Array(26).fill(0);
let currentChar1, currentChar2, currentIdx;
const charCodeA = "a".charCodeAt(0);
for (let idx = 0; idx < str1.length; idx++) {
currentChar1 = str1.charAt(idx);
currentIdx = currentChar1.charCodeAt(0) - charCodeA;
counter[currentIdx]++;
}
// Decrement the previously calculated counters
// by scanning through letters of str2.
for (let idx = 0; idx < str2.length; idx++) {
currentChar2 = str2.charAt(idx);
currentIdx = currentChar2.charCodeAt(0) - charCodeA;
console.log(currentIdx);
counter[currentIdx]--;
if (counter[currentIdx] < 0) {
return false;
}
}
return true;
}
console.log(isAnagram("rat", "car")); // true
function compare (a, b) {
var y = a.split("").sort().join(""),
z = b.split("").sort().join("");
console.log(z === y
? a + " and " + b + " are anagrams!"
: a + " and " + b + " are not anagrams."
);
}