feat: simpler solution for permutations (credit @JoshDevHub)

Author: nik-rev

16_permutations/solution/permutations-solution.js

@@ -1,27 +1,24 @@
-const permutations = function (
-  original,
-  partialPermutations = original.map((num) => [num]),
-) {
-  if (original.length <= 1) return [original];
+const permutations = function (array, index = 0, results = []) {
+  if (index == array.length) {
+    // We have formed a valid permutation.
 
-  const newPartialPermutations = [];
-  partialPermutations.forEach((partialPermutation) => {
-    const missingNums = original.filter(
-      (num) => !partialPermutation.includes(num),
-    );
-    missingNums.forEach((missingNum) =>
-      newPartialPermutations.push([...partialPermutation, missingNum]),
-    );
-  });
-
-  // We can pick any valid index because all of the elements will be the same length
-  const ANY_INDEX = 0;
+    // the [...array] syntax is a way to clone the contents of the array.
+    // because we do not want to pass a reference to the array, as that would mean
+    // that each item in `results` will be the same item
+    results.push([...array]);
+    return results;
+  }
 
-  if (newPartialPermutations[ANY_INDEX].length === original.length) {
-    return newPartialPermutations;
+  for (let i = index; i < array.length; i++) {
+    // We use "destructuring assignment" here to swap the values of array[index] and array[i]
+    //
+    // More info: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Destructuring_assignment
+    [array[index], array[i]] = [array[i], array[index]];
+    permutations(array, index + 1, results);
+    [array[index], array[i]] = [array[i], array[index]];
   }
 
-  return permutations(original, newPartialPermutations);
+  return results;
 };
 
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