Replace object with Map to store frequencies in 1.2 solution

Author: rroque98

JavaScript/chapter01/1.2 - Check Perm/rroque98_sol.js

@@ -1,36 +1,34 @@
 /*Given two strings, write a method to decide
 if one is apermutation of the other permutation:*/
 
-const isPermutation = (str1, str2) =>{
+const isPermutation = (str1, str2) => {
   if (str1.length !== str2.length) {
     return false;
   }
-  const obj1 = determineCharCount(str1);
-  const obj2 = determineCharCount(str2);
-  for (let char of str1) {
-    if (obj1[char] !== obj2[char]) {
+  const freq1 = determineCharCount(str1);
+  const freq2 = determineCharCount(str2);
+  for (const [char, count] of freq1) {
+    if (count !== freq2.get(char)) {
       return false;
     }
   }
   return true;
   //******** helper function ******
   function determineCharCount(string) {
-    return string.split('').reduce((acc, char) => {
-      if (acc[char] !== undefined) {
-        acc[char]++;
-      } else {
-        acc[char] = 1;
-      }
-      return acc;
-    }, {});
+    return string
+      .split('')
+      .reduce(
+        (acc, char) => acc.set(char, (acc.get(char) || 0) + 1),
+        new Map()
+      );
   }
-}
+};
 
 // Tests:
-  console.log(isPermutation('abc', 'abb') === false);
-  console.log(isPermutation('abb', 'abc') === false);
-  console.log(isPermutation('aaa', 'abc') === false);
-  console.log(isPermutation('abc', 'abcd') === false);
-  console.log(isPermutation('abc', 'bac') === true);
-  console.log(isPermutation('', '') === true);
-  console.log(isPermutation('12', '21') === true);
+console.log(isPermutation('abc', 'abb') === false);
+console.log(isPermutation('abb', 'abc') === false);
+console.log(isPermutation('aaa', 'abc') === false);
+console.log(isPermutation('abc', 'abcd') === false);
+console.log(isPermutation('abc', 'bac') === true);
+console.log(isPermutation('', '') === true);
+console.log(isPermutation('12', '21') === true);