A Number After A Double Reversal - Leetcode Solution

The problem A Number After a Double Reversal on LeetCode asks us to take an integer array as an input, perform two types of reversal, and return the final array after these operations. Let's understand the problem and break it down into steps.

Problem Statement

We are given an integer array nums. We need to apply two types of operations on this array, given below:

1. Reverse the elements of the subarray nums[0:k] (inclusive) and nums[k:len(nums)] (inclusive), where k is a given index from 1 to len(nums)-1.

2. Reverse the entire array nums.

We need to perform these operations in order and return the final array.

Solution

To solve this problem, we need to break it down into small steps and devise a plan accordingly. Let's look at the steps we can follow.

1. Reverse the subarray from nums[0:k] and nums[k:len(nums)].
2. Reverse the entire array nums.
3. Return the final array.

Step 1: Reverse the subarray

We need to reverse the subarray nums[0:k] and nums[k:len(nums)]. We can do this by swapping the values of the left and right indices until they meet in the middle. Let's write a function to do this:

def reverse_subarray(nums, start, end):
    while start < end:
        nums[start], nums[end] = nums[end], nums[start]
        start += 1
        end -= 1

This function takes an array nums, the starting index start and the end index end of the subarray that needs to be reversed. We swap the values of the left and right indices until they meet in the middle.

Step 2: Reverse the entire array

We need to reverse the entire array nums. We can use the same function we wrote in Step 1 to reverse the subarray nums[0:len(nums)]. Let's write a function to do this:

def reverse_array(nums):
    reverse_subarray(nums, 0, len(nums)-1)

This function takes an array nums and calls the reverse_subarray function with the starting index 0 and the end index len(nums)-1 to reverse the entire array.

Step 3: Return the final array

Now that we have reversed the subarray and the entire array, we need to return the final array as per the problem statement. Let's combine the above two functions to get the final solution:

def reverse_subarray(nums, start, end):
    while start < end:
        nums[start], nums[end] = nums[end], nums[start]
        start += 1
        end -= 1

def reverse_array(nums):
    reverse_subarray(nums, 0, len(nums)-1)

def reverse(nums, k):
    reverse_subarray(nums, 0, k)
    reverse_subarray(nums, k+1, len(nums)-1)
    reverse_array(nums)
    return nums

This function takes an array nums and an index k. We first reverse the subarray nums[0:k] and nums[k:len(nums)] using the reverse_subarray function. Then, we reverse the entire array nums using the reverse_array function. Finally, we return the final array.

Complexity Analysis

• Time Complexity: O(n) - We perform three passes on the array nums. The time complexity of the reverse_subarray function is O(n/2), and we call it twice. The time complexity of the reverse_array function is also O(n/2).

• Space Complexity: O(1) - We are not using any extra space and performing the operations in-place.

Conclusion

In this problem solution, we learned to perform array reversal and used the same concept to solve the given problem of A Number After a Double Reversal on LeetCode. We broke down the problem into small steps and followed a systematic approach to solve this problem.

1