We say that two strings are anagrams of each other if they have the exact same letters in same quantity of existance of each letter, regardless of non-alphabetic letters and case sensitivity.
Here's an example
// 'hello', 'elloh' -> anagrams
// 'love', 'hate' -> not anagrams
// 'i'm 26 years old' -> 'myeald oirs o' -> anagrams
Well, the Solution is Simple
We just need to remove all non-alphabetic letters first, and turn all letters into lower case.
// 'Hello World!@# ---30..' -> 'helloworld30'
const inputString = 'Hello World!@# ---30..'
inputString.toLowerCase().replace(/[\W_]+/g, ''); // returns 'helloworld30'
\W leaves the underscore, A short equivalent for [^a-zA-Z0-9] would be [\W_]
Then we need to convert string to array, sort the array alphabetically, and then turn it back into a string
// 'hello' -> ['h', 'e', 'l', 'l', 'o'] -> ['e', 'h', 'l', 'l', 'o'] -> ehllo
const inputString = 'Hello World!@# ---30..'
inputString.toLowerCase().replace(/[\W_]+/g, '').split('').sort().join(''); // returns '03dehllloorw'
Here's the final code
const anagrams = (firstInput, secondInput) => {
return (
firstInput
.toLowerCase()
.replace(/[\W_]+/g, '')
.split('')
.sort()
.join('') ===
secondInput
.toLowerCase()
.replace(/[\W_]+/g, '')
.split('')
.sort()
.join('')
);
}
Big O Notation
Time Complexity: O(n * Log n) because we've used sort algorithm
However, a Better solutions do exist, We'll also write another solution
const anagrams = (firstInput, secondInput) => {
firstInput = firstInput.toLowerCase().replace(/[\W_]+/g, '');
secondInput = secondInput.toLowerCase().replace(/[\W_]+/g, '');
if (firstInput.length !== secondInput.length) {
return false;
}
const inputLetterCount = {};
for (let i = 0; i < firstInput.length; i++) {
const currentLetter = firstInput[i];
inputLetterCount[currentLetter] = inputLetterCount[currentLetter] + 1 || 1;
}
for (let i = 0; i < secondInput.length; i++) {
const currentLetter = secondInput[i];
if (!inputLetterCount[currentLetter]) return false;
else inputLetterCount[currentLetter]--;
}
return true;
};
Big O Notation
Time Complexity: O(n)
Space Complexity: O(1)
Happy Coding ❤
Top comments (1)
you can also count the letters in a fixed length array. for e.g:
`
let d = "hello";
let k = "holle";
const arr = new Array(26).fill(0);
const arr2 = new Array(26).fill(0);
for(let i = 0; i < d.length; i++) {
arr[d[i].charCodeAt(0) - 'a'.charCodeAt(0)] += 1;
arr2[k[i].charCodeAt(0) - 'a'.charCodeAt(0)] += 1;
}
return arr.join(",") === arr2.join(",")
`