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Hash Map Approach in JavaScript

This post explores solving the

Solving “Determine if Two Strings Are Close” Using Hash Map in JavaScript

In algorithmic challenges like “Determine if Two Strings Are Close”, Hash Map provide an excellent solution due to their efficient data handling and retrieval capabilities.

Why Use Hash Map for String Comparison?

  • Efficient Character Counting: Hash Map allow for quick counting and comparison of character frequencies in strings.
  • Easy Comparison of Keys: Hash Map facilitate easy comparison of character presence across two strings.
  • Optimized Performance: They offer O(1) time complexity for character lookups and insertions, making the solution more efficient.

Understanding the Basics:

The “Determine if Two Strings Are Close” problem can be approached by comparing the frequency of each character in both strings and ensuring they can be transformed into each other.

Example: Determine if Two Strings Are Close

Problem Statement:

Determine if two strings are close. Two strings are close if you can attain one from the other using operations like rearranging characters or replacing one character with another character of the same frequency.

Solution Using Hash Map:

function areStringsClose(str1, str2) {
  if (str1.length !== str2.length) return false;
  const frequencyMap1 = new Map();
  const frequencyMap2 = new Map();

  for (let char of str1) {
    frequencyMap1.set(char, (frequencyMap1.get(char) || 0) + 1);

  for (let char of str2) {
    frequencyMap2.set(char, (frequencyMap2.get(char) || 0) + 1);

  if (new Set(frequencyMap1.values()).size !== new Set(frequencyMap2.values()).size) {
    return false;

  for (let [key, value] of frequencyMap1) {
    if (!frequencyMap2.has(key) || frequencyMap2.get(key) !== value) {
      return false;

  return true;

How the Code Works:

  • Frequency Counting: The function counts the frequency of each character in both strings using two separate Hash Maps.
  • Length Check: Initially, it checks if the lengths of the two strings are equal. If not, they can’t be close.
  • Comparing Frequency Sets: It then compares the sets of frequencies from both maps. If the sets differ in size, the strings aren’t close.
  • Character Frequency Match: Finally, it checks if each character’s frequency in one string matches its frequency in the other string.

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