Generate All Permutations of Array in DSA Javascript - Time & Space Complexity

Time Complexity: Generate All Permutations of Array

O(n!)

Understanding Time Complexity

We want to understand how the time needed to list all possible orders of an array grows as the array gets bigger.

How does the work increase when we add more items to the array?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function permute(arr) {
  const result = [];
  function backtrack(start = 0) {
    if (start === arr.length) {
      result.push(arr.slice());
      return;
    }
    for (let i = start; i < arr.length; i++) {
      [arr[start], arr[i]] = [arr[i], arr[start]];
      backtrack(start + 1);
      [arr[start], arr[i]] = [arr[i], arr[start]];
    }
  }
  backtrack();
  return result;
}
    

This code generates all possible orders (permutations) of the input array by swapping elements and exploring every option.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls combined with a loop that swaps elements to create new orders.
• How many times: The recursion explores every possible order, which grows very fast as the array size increases.

How Execution Grows With Input

Each time we add one more item, the number of orders multiplies by that new number.

Input Size (n)	Approx. Operations
3	6 (3 x 2 x 1)
5	120 (5 x 4 x 3 x 2 x 1)
7	5,040 (7 x 6 x 5 x 4 x 3 x 2 x 1)

Pattern observation: The work grows by multiplying all numbers from 1 up to n, which is much faster than just doubling or squaring.

Final Time Complexity

Time Complexity: O(n!)

This means the time needed grows extremely fast as the array gets bigger, because we must list every possible order.

Common Mistake

[X] Wrong: "The time grows like n squared because of the nested loop inside recursion."

[OK] Correct: The recursion multiplies the work by n at each level, creating factorial growth, which is much faster than just squaring.

Interview Connect

Understanding this growth helps you explain why generating all permutations is costly and when it's practical to use in real problems.

Self-Check

"What if we only wanted permutations of length k from n items? How would the time complexity change?"

Practice

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

easy

A. Finding the largest element in the array

B. Listing all possible orders of the array elements

C. Removing duplicate elements from the array

D. Sorting the array in ascending order

Solution

1. Step 1: Understand permutations

   Permutations are all possible ways to arrange elements in order.

2. Step 2: Apply to array elements

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

3. Final Answer:

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

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
• Believing permutations find max element

2. Which of the following is the correct syntax to start a recursive function for generating permutations in JavaScript?

easy

A. function permute(arr l r) { }

B. function permute(arr, l, r) => { }

C. permute function(arr, l, r) { }

D. function permute(arr, l, r) { }

Solution

1. Step 1: Recall JavaScript function syntax

   Functions are declared with the keyword 'function', parameters inside parentheses separated by commas.

2. Step 2: Check each option

   function permute(arr, l, r) { } uses correct syntax: function name, parameters with commas, and braces. function permute(arr, l, r) => { } uses arrow syntax incorrectly. permute function(arr, l, r) { } has wrong order. function permute(arr l r) { } misses commas.

3. Final Answer:

   function permute(arr, l, r) { } -> Option D

4. Quick Check:

   Correct function syntax = function permute(arr, l, r) { } [OK]

Hint: Use 'function name(params) { }' for recursive functions [OK]

Common Mistakes:

• Using arrow syntax without proper parameters
• Missing commas between parameters
• Incorrect function keyword placement

3. What is the output of the following code?

function permute(arr, l, r) {
  if (l === r) {
    console.log(arr.join(''));
  } else {
    for (let i = l; i <= r; i++) {
      [arr[l], arr[i]] = [arr[i], arr[l]];
      permute(arr, l + 1, r);
      [arr[l], arr[i]] = [arr[i], arr[l]];
    }
  }
}
permute(['A', 'B', 'C'], 0, 2);

medium

A. ABC, ACB, BAC, BCA, CAB, CBA

B. ABC, BAC, CAB

C. ABC only

D. SyntaxError

Solution

1. Step 1: Understand the recursive swapping

   The function swaps elements starting from index l to r, generating all permutations by recursive calls.

2. Step 2: Trace output for ['A','B','C']

   All 6 permutations of 3 elements are printed: ABC, ACB, BAC, BCA, CAB, CBA.

3. Final Answer:

   ABC, ACB, BAC, BCA, CAB, CBA -> Option A

4. Quick Check:

   All 3-element permutations = ABC, ACB, BAC, BCA, CAB, CBA [OK]

Hint: Swapping + recursion = all permutations [OK]

Common Mistakes:

• Missing swap back step causing wrong output
• Printing partial permutations only
• Confusing output order

4. Identify the error in this permutation function:

function permute(arr, l, r) {
  if (l === r) {
    console.log(arr.join(''));
  } else {
    for (let i = l; i < r; i++) {
      [arr[l], arr[i]] = [arr[i], arr[l]];
      permute(arr, l + 1, r);
      [arr[l], arr[i]] = [arr[i], arr[l]];
    }
  }
}
permute(['X', 'Y', 'Z'], 0, 2);

medium

A. The for loop should use i <= r instead of i < r

B. The swap syntax is incorrect

C. The base case condition should be l > r

D. The function should not use recursion

Solution

1. Step 1: Check loop boundary

   The loop runs from i = l to i < r, so it misses the last index r.

2. Step 2: Correct loop condition

   Changing to i <= r includes the last element, ensuring all permutations are generated.

3. Final Answer:

   The for loop should use i <= r instead of i < r -> Option A

4. Quick Check:

   Loop boundary must include r [OK]

Hint: Loop to r inclusive for full permutations [OK]

Common Mistakes:

• Using i < r misses last element
• Incorrect swap syntax causing errors
• Wrong base case stopping recursion early

5. Given an array with duplicate elements ['A', 'B', 'A'], how can you modify the permutation function to avoid printing duplicate permutations?

hard

A. Use a set to track and skip already generated permutations

B. Remove duplicates from the array before permutation

C. Sort the array and skip swapping identical elements at the same recursion level

D. No modification needed; duplicates are automatically avoided

Solution

1. Step 1: Understand duplicates in permutations

   Duplicates cause repeated permutations when identical elements are swapped.

2. Step 2: Skip swapping identical elements

   Sorting the array and checking if the element was already swapped at the current recursion level avoids duplicates.

3. Final Answer:

   Sort the array and skip swapping identical elements at the same recursion level -> Option C

4. Quick Check:

   Skip identical swaps after sorting to avoid duplicates [OK]

Hint: Sort + skip identical swaps to avoid duplicate permutations [OK]

Common Mistakes:

• Removing duplicates loses valid permutations
• Using set only after generating all permutations (inefficient)
• Assuming duplicates are avoided automatically