Generate All Permutations of Array in DSA Typescript

Mental Model

We want to list every possible order of the items in an array by swapping elements step by step.

Analogy: Imagine you have 3 different colored balls and want to see all ways to arrange them in a row by swapping their positions.

Array: [1, 2, 3]
Indexes:  0  1  2
Start with first index and swap with each possible index to create new orders.

Dry Run Walkthrough

Input: array: [1, 2, 3]

Goal: Generate all possible orders of the array elements

Step 1: Start with index 0, swap element at 0 with 0 (no change)

[1, 2, 3]

Why: Begin permutations by fixing first element and permuting the rest

Step 2: Move to index 1, swap element at 1 with 1 (no change)

[1, 2, 3]

Why: Fix second element and permute remaining elements

Step 3: Move to index 2, reached end, record permutation [1, 2, 3]

[1, 2, 3]

Why: Base case: one permutation completed

Step 4: Backtrack to index 1, swap element at 1 with 2

[1, 3, 2]

Why: Swap to create new permutation branch

Step 5: Move to index 2, reached end, record permutation [1, 3, 2]

[1, 3, 2]

Why: Base case: another permutation completed

Step 6: Backtrack to index 0, swap element at 0 with 1

[2, 1, 3]

Why: Change first element to generate new permutations

Step 7: At index 1, swap element at 1 with 1 (no change)

[2, 1, 3]

Why: Fix second element and permute rest

Step 8: At index 2, reached end, record permutation [2, 1, 3]

[2, 1, 3]

Why: Record new permutation

Result:

[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]

Annotated Code

DSA Typescript

class Permutations {
  private results: number[][] = [];

  public generate(nums: number[]): number[][] {
    this.backtrack(nums, 0);
    return this.results;
  }

  private backtrack(nums: number[], start: number): void {
    if (start === nums.length) {
      this.results.push([...nums]); // record current permutation
      return;
    }

    for (let i = start; i < nums.length; i++) {
      [nums[start], nums[i]] = [nums[i], nums[start]]; // swap to fix element at start
      this.backtrack(nums, start + 1); // permute remaining elements
      [nums[start], nums[i]] = [nums[i], nums[start]]; // swap back to restore
    }
  }
}

const perm = new Permutations();
const input = [1, 2, 3];
const output = perm.generate(input);
for (const arr of output) {
  console.log(arr.join(", "));
}

if (start === nums.length) {

Base case: all positions fixed, record current permutation

this.results.push([...nums]);

Save a copy of current permutation

[nums[start], nums[i]] = [nums[i], nums[start]];

Swap to fix element at current start index

this.backtrack(nums, start + 1);

Recurse to fix next position

[nums[start], nums[i]] = [nums[i], nums[start]];

Swap back to restore original array for next iteration

Output

1, 2, 3

1, 3, 2

2, 1, 3

2, 3, 1

3, 1, 2

3, 2, 1

Complexity Analysis

Time: O(n! * n) because there are n! permutations and copying each permutation takes O(n)

Space: O(n) for recursion stack plus O(n! * n) for storing all permutations

vs Alternative: Compared to generating permutations by building new arrays each time, swapping in place reduces extra space but still requires factorial time due to permutation count

Edge Cases

empty array []

returns [[]] as the only permutation (empty permutation)

DSA Typescript

if (start === nums.length) {

array with one element [5]

returns [[5]] as the only permutation

DSA Typescript

if (start === nums.length) {

array with duplicate elements [1, 1]

returns all permutations including duplicates, e.g. [1,1] and [1,1]

DSA Typescript

for (let i = start; i < nums.length; i++) {

When to Use This Pattern

When asked to list all possible orders or arrangements of items, use backtracking with swapping to generate permutations efficiently.

Common Mistakes

Mistake: Not swapping back after recursive call, causing incorrect array state
Fix: Always swap back after recursion to restore original order for next iteration

Mistake: Pushing reference to nums instead of a copy, causing all results to be the same
Fix: Push a copy of the array using spread operator or slice

Summary

Generates all possible orders of an array by swapping elements recursively.

Use when you need every arrangement of items, like permutations in puzzles or combinations.

The key is to fix one element at a time by swapping and backtracking to explore all options.

Practice

1. What does the process of generating all permutations of an array mean?

easy

A. Listing all possible orders of the array elements

B. Sorting the array in ascending order

C. Removing duplicate elements from the array

D. Finding the largest element in the array

Solution

1. Step 1: Understand permutations

   Permutations are all possible ways to arrange elements in order.

2. Step 2: Apply to array

   Generating all permutations means listing every possible order of the array elements.

3. Final Answer:

   Listing all possible orders of the array elements -> Option A

4. Quick Check:

   Permutations = all orders [OK]

Hint: Permutations = all ways to arrange elements [OK]

Common Mistakes:

• Confusing permutations with sorting
• Thinking permutations remove duplicates
• Mixing permutations with finding max element

2. Which of the following TypeScript function signatures correctly represents a function that generates all permutations of an array of numbers?

easy

A. function permute(arr: number[][]): number[]

B. function permute(arr: number): number[]

C. function permute(arr: number[]): number[][]

D. function permute(arr: string[]): string

Solution

1. Step 1: Analyze input and output types

   The input is an array of numbers, so type is number[]. The output is all permutations, which is an array of arrays of numbers, so number[][].

2. Step 2: Match function signature

   function permute(arr: number[]): number[][] matches input number[] and output number[][] correctly.

3. Final Answer:

   function permute(arr: number[]): number[][] -> Option C

4. Quick Check:

   Input array, output array of arrays [OK]

Hint: Input array, output array of arrays for permutations [OK]

Common Mistakes:

• Using single number instead of array as input
• Returning a single array instead of array of arrays
• Mixing string and number types incorrectly

3. What is the output of the following TypeScript code?

function permute(arr: number[]): number[][] {
  if (arr.length === 0) return [[]];
  const result: number[][] = [];
  for (let i = 0; i < arr.length; i++) {
    const rest = [...arr.slice(0, i), ...arr.slice(i + 1)];
    for (const perm of permute(rest)) {
      result.push([arr[i], ...perm]);
    }
  }
  return result;
}

console.log(permute([1, 2]));

medium

A. [[1, 2], [2, 1]]

B. [[1, 2]]

C. [[2, 1]]

D. []

Solution

1. Step 1: Understand base case and recursion

   Base case returns [[]] for empty array. For [1,2], it picks each element and permutes the rest.

2. Step 2: Trace recursive calls for [1,2]

   First i=0: arr[i]=1, rest=[2], permute([2]) returns [[2]], so push [1,2]. Then i=1: arr[i]=2, rest=[1], permute([1]) returns [[1]], so push [2,1].

3. Final Answer:

   [[1, 2], [2, 1]] -> Option A

4. Quick Check:

   Permutations of [1,2] = [[1,2],[2,1]] [OK]

Hint: Check base case and recursive build-up [OK]

Common Mistakes:

• Forgetting to include all permutations
• Returning only one permutation
• Confusing order of elements

4. Identify the error in this TypeScript function that attempts to generate all permutations of an array:

function permute(arr: number[]): number[][] {
  if (arr.length === 1) return [arr];
  const result: number[][] = [];
  for (let i = 0; i < arr.length; i++) {
    const rest = arr.slice(1);
    for (const perm of permute(rest)) {
      result.push([arr[i], ...perm]);
    }
  }
  return result;
}

medium

A. The base case should return an empty array

B. The function should return a flat array, not nested arrays

C. The loop should start from 1 instead of 0

D. The slice should exclude the current element, not always the first

Solution

1. Step 1: Analyze the slice operation

   The code uses arr.slice(1) always, which removes only the first element, not the current element at index i.

2. Step 2: Correct slice to exclude current element

   To generate permutations correctly, slice should exclude arr[i], e.g., [...arr.slice(0, i), ...arr.slice(i + 1)].

3. Final Answer:

   The slice should exclude the current element, not always the first -> Option D

4. Quick Check:

   Exclude current element in slice [OK]

Hint: Slice must exclude current index element [OK]

Common Mistakes:

• Always slicing from index 1 ignoring i
• Wrong base case length check
• Returning wrong data structure

5. Given an array with possible duplicate elements, how can you modify the permutation generation function to avoid duplicate permutations?

function permuteUnique(arr: number[]): number[][] {
  // Your code here
}

hard

A. Use a Set to store all permutations after generation

B. Sort the array first and skip duplicates during recursion

C. Remove duplicates from the array before generating permutations

D. Generate permutations normally without any change

Solution

1. Step 1: Understand duplicate permutations problem

   Duplicates in input cause repeated permutations if not handled.

2. Step 2: Sort and skip duplicates during recursion

   Sorting helps group duplicates. During recursion, skip elements equal to previous to avoid repeated branches.

3. Step 3: Why other options fail

   Using Set after generation is costly. Removing duplicates loses valid permutations. No change causes duplicates.

4. Final Answer:

   Sort the array first and skip duplicates during recursion -> Option B

5. Quick Check:

   Sort + skip duplicates = unique permutations [OK]

Hint: Sort array and skip duplicates in recursion [OK]

Common Mistakes:

• Removing duplicates from input loses valid permutations
• Using Set after generation wastes time and memory
• Ignoring duplicates causes repeated permutations