Generate All Permutations of Array in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the permutation generation by calling the helper function.

DSA Typescript

function permute(nums: number[]): number[][] {
  const result: number[][] = [];
  function backtrack(start: number) {
    // helper function body
  }
  [1];
  return result;
}

Drag options to blanks, or click blank then click option'

Abacktrack(0)

Bbacktrack(nums.length)

Cbacktrack(-1)

Dbacktrack()

Attempts:

3 left

💡 Hint

Common Mistakes

Calling backtrack with the length of the array instead of 0.

Calling backtrack without any argument.

Starting from a negative index.

We start generating permutations from index 0, so calling backtrack(0) initiates the process correctly.

2fill in blank

medium

Complete the code to swap elements at two indices in the array.

DSA Typescript

function swap(nums: number[], i: number, j: number): void {
  const temp = nums[i];
  nums[i] = nums[[1]];
  nums[[1]] = temp;
}

Drag options to blanks, or click blank then click option'

A0

Bj

Cnums.length - 1

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using i instead of j for the second index.

Using a fixed index like 0 or nums.length - 1.

To swap elements at indices i and j, we assign nums[i] = nums[j] and nums[j] = temp.

3fill in blank

hard

Fix the error in the base case condition to add a copy of the current permutation to the result.

DSA Typescript

function permute(nums: number[]): number[][] {
  const result: number[][] = [];
  function backtrack(start: number) {
    if (start [1] nums.length) {
      result.push([...nums]);
      return;
    }
    // rest of code
  }
  backtrack(0);
  return result;
}

Drag options to blanks, or click blank then click option'

A>

B<

C===

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using less than or greater than instead of equality.

Using less than or equal which triggers too early.

The base case is when start equals nums.length, meaning a full permutation is formed.

4fill in blank

hard

Fill both blanks to complete the for loop and swap calls inside the backtrack function.

DSA Typescript

function backtrack(start: number) {
  if (start === nums.length) {
    result.push([...nums]);
    return;
  }
  for (let i = [1]; i < nums.length; i++) {
    swap(nums, start, i);
    backtrack([2]);
    swap(nums, start, i);
  }
}

Drag options to blanks, or click blank then click option'

Astart

Bi

Cstart + 1

Di + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop from 0 instead of start.

Calling backtrack with i instead of start + 1.

The loop starts from 'start' index, and after swapping, we recurse with start + 1 to fix the next position.

5fill in blank

hard

Fill all three blanks to complete the full permutation function with swap helper.

DSA Typescript

function permute(nums: number[]): number[][] {
  const result: number[][] = [];
  function swap(arr: number[], i: number, j: number): void {
    const temp = arr[i];
    arr[i] = arr[[1]];
    arr[[2]] = temp;
  }
  function backtrack(start: number) {
    if (start === nums.length) {
      result.push([...nums]);
      return;
    }
    for (let i = [3]; i < nums.length; i++) {
      swap(nums, start, i);
      backtrack(start + 1);
      swap(nums, start, i);
    }
  }
  backtrack(0);
  return result;
}

Drag options to blanks, or click blank then click option'

Aj

Bi

Cstart

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up i and j in swap assignments.

Starting loop from 0 instead of start.

In swap, we assign arr[i] = arr[j] and arr[j] = temp. The loop in backtrack starts from 'start' index.

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