Next Permutation of Array in DSA Go - Time & Space Complexity

Time Complexity: Next Permutation of Array

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the next permutation changes as the array size grows.

How does the number of steps grow when the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func nextPermutation(nums []int) {
    n := len(nums)
    i := n - 2
    for i >= 0 && nums[i] >= nums[i+1] {
        i--
    }
    if i >= 0 {
        j := n - 1
        for nums[j] <= nums[i] {
            j--
        }
        nums[i], nums[j] = nums[j], nums[i]
    }
    reverse(nums[i+1:])
}

func reverse(nums []int) {
    left, right := 0, len(nums)-1
    for left < right {
        nums[left], nums[right] = nums[right], nums[left]
        left++
        right--
    }
}
    

This code finds the next lexicographical permutation of the array in place.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Scanning the array from the end to find the first decreasing element, then scanning again to find the element to swap, and finally reversing a subarray.
• How many times: Each loop can run up to n times in the worst case, where n is the array length.

How Execution Grows With Input

As the array size grows, the number of steps grows roughly in a straight line with n.

Input Size (n)	Approx. Operations
10	Up to about 30 steps
100	Up to about 300 steps
1000	Up to about 3000 steps

Pattern observation: The operations grow linearly with input size, meaning doubling the array roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the next permutation grows in direct proportion to the array size.

Common Mistake

[X] Wrong: "The algorithm must check all permutations, so it takes factorial time."

[OK] Correct: The algorithm only scans the array a few times, never generating all permutations, so it runs much faster than factorial time.

Interview Connect

Understanding this linear time approach shows you can handle array manipulations efficiently, a skill often needed in coding challenges and interviews.

Self-Check

"What if the array was sorted in descending order? How would the time complexity change?"

Practice

1. What does the next permutation of an array represent?

easy

A. The next lexicographically greater arrangement of the array elements

B. The array sorted in ascending order

C. The array reversed

D. The smallest element moved to the front

Solution

1. Step 1: Understand the concept of permutation order

   Permutations can be ordered lexicographically like words in a dictionary.

2. Step 2: Define next permutation

   The next permutation is the next arrangement that is just larger than the current one in this order.

3. Final Answer:

   The next lexicographically greater arrangement of the array elements -> Option A

4. Quick Check:

   Next permutation = next bigger arrangement [OK]

Hint: Think of next permutation as next dictionary word [OK]

Common Mistakes:

• Confusing next permutation with sorting
• Thinking it always reverses the array
• Assuming it moves smallest element to front

2. Which Go code snippet correctly swaps two elements nums[i] and nums[j] in an array?

easy

A. swap(nums[i], nums[j])

B. temp = nums[i]; nums[i] = nums[j]; nums[j] = temp

C. nums[i], nums[j] = nums[j], nums[i]

D. nums[i] = nums[j]; nums[j] = nums[i]

Solution

1. Step 1: Recall Go multiple assignment syntax

   Go allows swapping two variables in one line using multiple assignment.

2. Step 2: Identify correct swap syntax

   The syntax nums[i], nums[j] = nums[j], nums[i] swaps values correctly without temporary variable.

3. Final Answer:

   nums[i], nums[j] = nums[j], nums[i] -> Option C

4. Quick Check:

   Go swap uses multiple assignment [OK]

Hint: Use Go's multiple assignment for swapping [OK]

Common Mistakes:

• Using undefined swap function
• Forgetting temporary variable in manual swap
• Assigning both to same value causing loss

3. What is the output of the following Go code that finds the next permutation of nums := []int{1, 3, 2}?

func nextPermutation(nums []int) {
    i := len(nums) - 2
    for i >= 0 && nums[i] >= nums[i+1] {
        i--
    }
    if i >= 0 {
        j := len(nums) - 1
        for nums[j] <= nums[i] {
            j--
        }
        nums[i], nums[j] = nums[j], nums[i]
    }
    left, right := i+1, len(nums)-1
    for left < right {
        nums[left], nums[right] = nums[right], nums[left]
        left++
        right--
    }
}

func main() {
    nums := []int{1, 3, 2}
    nextPermutation(nums)
    fmt.Println(nums)
}

medium

A. [1, 2, 3]

B. [2, 1, 3]

C. [1, 3, 2]

D. [3, 1, 2]

Solution

1. Step 1: Find first decreasing element from right

   Starting from index 1 (value 3), check if nums[i] >= nums[i+1]. At i=1, 3 >= 2 true, i=0, 1 >= 3 false, so i=0.

2. Step 2: Find element just larger than nums[i]

   From right, find j where nums[j] > nums[i] (nums[0]=1). j=2 (value 2) > 1, swap nums[0] and nums[2]. Array becomes [2,3,1]. Reverse suffix from i+1=1 to end: reverse [3,1] to [1,3]. Final array: [2,1,3].

3. Final Answer:

   [2, 1, 3] -> Option B

4. Quick Check:

   Next permutation of [1,3,2] = [2,1,3] [OK]

Hint: Swap and reverse suffix after first decreasing index [OK]

Common Mistakes:

• Not reversing suffix after swap
• Swapping wrong elements
• Stopping at first decreasing element without swap

4. Identify the bug in this Go code snippet for next permutation:

func nextPermutation(nums []int) {
    i := len(nums) - 1
    for i > 0 && nums[i] <= nums[i-1] {
        i--
    }
    if i > 0 {
        j := len(nums) - 1
        for nums[j] <= nums[i-1] {
            j--
        }
        nums[i-1], nums[j] = nums[j], nums[i-1]
    }
    left, right := i, len(nums)-1
    for left < right {
        nums[left], nums[right] = nums[right], nums[left]
        left++
        right--
    }
}

medium

A. The initial index i should start at len(nums) - 2, not len(nums) - 1

B. The swap should be between nums[i] and nums[j], not nums[i-1] and nums[j]

C. The reversal should start from i-1, not i

D. The loop conditions should use < instead of >

Solution

1. Step 1: Check starting index for finding first decreasing element

   The algorithm must start checking from second last element (len(nums)-2) to compare with next element.

2. Step 2: Identify impact of starting at len(nums)-1

   Starting at len(nums)-1 compares last element with out-of-range index, causing incorrect behavior or skipping needed checks.

3. Final Answer:

   The initial index i should start at len(nums) - 2, not len(nums) - 1 -> Option A

4. Quick Check:

   Start i at len(nums)-2 for correct next permutation [OK]

Hint: Start i at second last index to find decreasing pair [OK]

Common Mistakes:

• Starting i at last index causing index errors
• Swapping wrong elements
• Incorrect reversal range

5. Given an array nums := []int{3, 2, 1}, what will be the next permutation after applying the standard next permutation algorithm? Explain why.

hard

A. [3, 2, 1] because it is already the highest permutation

B. [3, 1, 2] because swapping first and last elements gives next permutation

C. [2, 3, 1] because reversing the array gives next permutation

D. [1, 2, 3] because the array is in descending order, so next permutation is the lowest ascending order

Solution

1. Step 1: Check if array is in descending order

   Array [3, 2, 1] is strictly descending, meaning it is the highest permutation possible.

2. Step 2: Apply next permutation rule for highest permutation

   When highest permutation is reached, next permutation is the lowest ascending order, which is [1, 2, 3].

3. Final Answer:

   [1, 2, 3] because the array is in descending order, so next permutation is the lowest ascending order -> Option D

4. Quick Check:

   Descending array next permutation = ascending array [OK]

Hint: If descending, next permutation is ascending sorted array [OK]

Common Mistakes:

• Thinking next permutation is same as current if descending
• Swapping wrong elements instead of reversing whole array
• Not resetting to lowest permutation after highest