Cracking the Coding Interview: Check Permutation

This post is part of the Algorithms Problem Solving series and a problem from the Cracking the Coding Interview book.

Problem description

Given two strings, write a method to decide if one is a permutation of the other


input: 'aba', 'aab'
output: true

input: 'aba', 'aaba'
output: false

input: 'aba', 'aa'
output: false


If they have different lengths, they are not permutations.

Build a hashmap of characters of one of the strings Iterate through the other string and check if the characters are in the hashmap If there's no match in the hashmap, they are not permutations If it passes the loop, they are permutations.

  • Runtime Complexity: O(N), where N = the length of the string
  • Space Complexity: O(N), where N = the length of the string
function buildCharsMap(string) {
  const charsMap = new Map();

  for (let char of string) {
    if (charsMap.has(char)) charsMap.set(char, charsMap.get(char) + 1);
    else charsMap.set(char, 1);

  return charsMap;

function checkPermutation(s1, s2) {
  if (s1.length !== s2.length) return false;
  const charsMap = buildCharsMap(s1);

  for (let char of s2) {
    if (!charsMap.has(char)) {
      return false;

    if (charsMap.get(char) === 0) {
      return false;

    charsMap.set(char, charsMap.get(char) - 1);

  return true;


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