Next Permutation

 

1. Clarify the problem: Before diving into solving the problem, let's clarify the requirements:

  • We are given an array of integers representing a permutation.
  • We need to find the next permutation in lexicographical order.
  • If no next permutation is possible, we need to return the array in ascending order.

2. Analyze the problem: To solve this problem, we need to understand how permutations work and how to generate the next permutation. We can find the next permutation by following these steps:

  1. Start from the rightmost element and find the first pair of adjacent elements nums[i] and nums[i-1] such that nums[i-1] < nums[i].
  2. Find the smallest element in the subarray nums[i:] that is greater than nums[i-1]. Let's call this element nums[j].
  3. Swap nums[i-1] with nums[j].
  4. Reverse the subarray nums[i:] to get the next lexicographical permutation.
  5. If no pair nums[i-1] < nums[i] is found in step 1, it means the array is already sorted in descending order. In this case, we need to reverse the entire array to get the next permutation.

3. Design an algorithm: Here is the algorithm to find the next permutation:

  1. Find the first pair nums[i-1] and nums[i] such that nums[i-1] < nums[i]. If no such pair exists, go to step 4.
  2. Find the smallest element nums[j] in the subarray nums[i:] that is greater than nums[i-1].
  3. Swap nums[i-1] with nums[j].
  4. Reverse the subarray nums[i:].
  5. Return the modified array.

4. Explain your approach: The approach involves finding the first pair of adjacent elements in the array such that the left element is smaller than the right element. This pair indicates the starting point for the next permutation. We then find the smallest element in the subarray to the right of nums[i-1] that is greater than nums[i-1]. We swap these two elements, and then reverse the subarray to the right of nums[i-1] to get the next lexicographical permutation. If no such pair is found, it means the array is already in descending order, so we reverse the entire array to get the next permutation.

5. Write clean and readable code:

python
def nextPermutation(nums):
    n = len(nums)
    i = n - 2

    # Find the first pair nums[i-1] < nums[i]
    while i >= 0 and nums[i] >= nums[i + 1]:
        i -= 1

    if i >= 0:
        # Find the smallest element nums[j] > nums[i-1]
        j = n - 1
        while j >= 0 and nums[j] <= nums[i]:
            j -= 1
        # Swap nums[i-1] and nums[j]
        nums[i], nums[j] = nums[j], nums[i]

    # Reverse the subarray nums[i:]
    left = i + 1
    right = n - 1
    while left < right:
        nums[left], nums[right] = nums[right], nums[left]
        left += 1
        right -= 1

    return nums

6. Test your code:

python
# Test case 1
nums1 = [1, 2, 3]
print(nextPermutation(nums1))  # [1, 3, 2]

# Test case 2
nums2 = [3, 2, 1]
print(nextPermutation(nums2))  # [1, 2, 3]

# Test case 3
nums3 = [1, 1, 5]
print(nextPermutation(nums3))  # [1, 5, 1]

# Test case 4
nums4 = [1, 3, 2]
print(nextPermutation(nums4))  # [2, 1, 3]

7. Optimize if necessary: The solution provided already follows an optimal approach with a time complexity of O(n), where n is the length of the input array. We traverse the array twice, once to find the pair nums[i-1] and nums[i] and once to reverse the subarray. The space complexity is O(1) as we are modifying the input array in place.

8. Handle error cases: The code assumes that the input array contains valid integers. However, it does not handle invalid input or error cases explicitly. If the input is not a valid permutation or contains non-integer elements, the behavior of the code may be unpredictable.

9. Discuss complexity analysis: The time complexity of the solution is O(n), where n is the length of the input array. This is because we traverse the array twice: once to find the pair nums[i-1] and nums[i] and once to reverse the subarray. The space complexity is O(1) as we are modifying the input array in place. There are no additional data structures used that grow with the input size.

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