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

Time Complexity: Generate All Permutations of Array

O(n!)

Understanding Time Complexity

We want to understand how the time needed grows when we generate all possible orders of an array.

How does the work increase as the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func permute(nums []int) [][]int {
    var res [][]int
    var backtrack func(start int)
    backtrack = func(start int) {
        if start == len(nums) {
            temp := make([]int, len(nums))
            copy(temp, nums)
            res = append(res, temp)
            return
        }
        for i := start; i < len(nums); i++ {
            nums[start], nums[i] = nums[i], nums[start]
            backtrack(start + 1)
            nums[start], nums[i] = nums[i], nums[start]
        }
    }
    backtrack(0)
    return res
}

This code generates all possible orders (permutations) of the input array by swapping elements recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls with a loop that swaps elements to create new permutations.
• How many times: The recursion explores every possible order, which is factorial in number of elements.

How Execution Grows With Input

Each time we add one more element, the number of orders multiplies by that 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	5040 (7 x 6 x 5 x 4 x 3 x 2 x 1)

Pattern observation: The work grows very fast, multiplying by the new size each time.

Final Time Complexity

Time Complexity: O(n!)

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

Common Mistake

[X] Wrong: "The time grows like n squared or n cubed because of the loops."

[OK] Correct: The recursion creates all possible orders, which is much more than simple loops; it grows factorially, not polynomially.

Interview Connect

Understanding this growth helps you explain why generating all permutations is expensive and when it is practical to use.

Self-Check

"What if we only wanted permutations of length k from n elements? 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. Sorting the array in ascending order

C. Listing all possible orders of the array elements

D. Removing duplicate elements from the array

Solution

1. Step 1: Understand permutations

   Permutations are all possible ways to arrange elements in order.

2. Step 2: Compare options

   Only Listing all possible orders of the array elements describes listing all possible orders of elements.

3. Final Answer:

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

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 Go code snippet correctly starts a recursive function to generate permutations of an integer slice arr?

easy

A. func permute(arr int[], start int) {}

B. func permute(arr []int, start int) {}

C. func permute(arr []int, start string) {}

D. func permute(arr []int) {}

Solution

1. Step 1: Check Go function syntax

   Go uses []int for slices and int for integers; parameters must match expected types.

2. Step 2: Validate parameters for permutation

   We need the array slice and an integer index to track recursion start.

3. Final Answer:

   func permute(arr []int, start int) {} -> Option B

4. Quick Check:

   Correct Go slice and int params [OK]

Hint: Use []int for slices and int for indices [OK]

Common Mistakes:

• Using incorrect slice syntax like int[]
• Using wrong parameter types like string for index
• Omitting necessary parameters for recursion

3. What will be the output of this Go code snippet that generates permutations of []int{1, 2} and prints them?

func permute(arr []int, start int) {
  if start == len(arr) {
    fmt.Println(arr)
    return
  }
  for i := start; i < len(arr); i++ {
    arr[start], arr[i] = arr[i], arr[start]
    permute(arr, start+1)
    arr[start], arr[i] = arr[i], arr[start]
  }
}

func main() {
  permute([]int{1, 2}, 0)
}

medium

A. Compilation error

B. [2 1]
[1 2]

C. [1 2]

D. [1 2]
[2 1]

Solution

1. Step 1: Trace recursion for arr = [1, 2]

   Start=0: swap arr[0] with arr[0] -> [1,2], recurse start=1; then swap arr[0] with arr[1] -> [2,1], recurse start=1.

2. Step 2: Print when start == len(arr)

   At start=2, print current arr. So prints [1 2] then [2 1].

3. Final Answer:

   [1 2] [2 1] -> Option D

4. Quick Check:

   All permutations printed in order [OK]

Hint: Swap and backtrack to print all permutations [OK]

Common Mistakes:

• Not swapping back after recursion
• Printing before full permutation formed
• Confusing order of printed permutations

4. Identify the bug in this Go function intended to generate all permutations of arr:

func permute(arr []int, start int) {
  if start == len(arr) {
    fmt.Println(arr)
    return
  }
  for i := 0; i < len(arr); i++ {
    arr[start], arr[i] = arr[i], arr[start]
    permute(arr, start+1)
    arr[start], arr[i] = arr[i], arr[start]
  }
}

medium

A. Loop should start from 'start' instead of 0

B. Missing base case for recursion

C. Swapping elements is incorrect

D. Function should return a slice, not print

Solution

1. Step 1: Analyze loop range

   The loop runs from i=0 to len(arr), but it should start from 'start' to avoid duplicate swaps.

2. Step 2: Understand impact of loop start

   Starting from 0 causes repeated permutations and incorrect swapping order.

3. Final Answer:

   Loop should start from 'start' instead of 0 -> Option A

4. Quick Check:

   Loop start index = start [OK]

Hint: Loop from start index to avoid duplicates [OK]

Common Mistakes:

• Looping from zero causing repeated permutations
• Not swapping back after recursion
• Ignoring base case correctness

5. Given an array []int{1, 2, 3}, how many unique permutations will be generated by a correct permutation function, and why?

hard

A. 6, because 3! (3 factorial) = 6 permutations

B. 3, because there are 3 elements

C. 9, because 3 * 3 = 9 combinations

D. 1, because the array is fixed

Solution

1. Step 1: Calculate permutations count for 3 elements

   Number of permutations = factorial of number of elements = 3! = 3*2*1 = 6.

2. Step 2: Understand why other options are incorrect

   3 is just count of elements, 9 is product not permutations, 1 is incorrect as array can reorder.

3. Final Answer:

   6, because 3! (3 factorial) = 6 permutations -> Option A

4. Quick Check:

   Permutations of 3 elements = 6 [OK]

Hint: Permutations count = factorial of array length [OK]

Common Mistakes:

• Confusing permutations count with array length
• Using multiplication instead of factorial
• Thinking only one arrangement exists