Introduction

In this blog post, we will explore how to solve the LeetCode problem "2460" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a 0-indexed array nums of size n consisting of non-negative integers. You need to apply n - 1 operations to this array where, in the ith operation (0-indexed), you will apply the following on the ith element of nums: If nums[i] == nums[i + 1], then multiply nums[i] by 2 and set nums[i + 1] to 0. Otherwise, you skip this operation. After performing all the operations, shift all the 0's to the end of the array. For example, the array [1,0,2,0,0,1] after shifting all its 0's to the end, is [1,2,1,0,0,0]. Return the resulting array. Note that the operations are applied sequentially, not all at once. Example 1: Input: nums = [1,2,2,1,1,0] Output: [1,4,2,0,0,0] Explanation: We do the following operations: - i = 0: nums[0] and nums[1] are not equal, so we skip this operation. - i = 1: nums[1] and nums[2] are equal, we multiply nums[1] by 2 and change nums[2] to 0. The array becomes [1,4,0,1,1,0]. - i = 2: nums[2] and nums[3] are not equal, so we skip this operation. - i = 3: nums[3] and nums[4] are equal, we multiply nums[3] by 2 and change nums[4] to 0. The array becomes [1,4,0,2,0,0]. - i = 4: nums[4] and nums[5] are equal, we multiply nums[4] by 2 and change nums[5] to 0. The array becomes [1,4,0,2,0,0]. After that, we shift the 0's to the end, which gives the array [1,4,2,0,0,0]. Example 2: Input: nums = [0,1] Output: [1,0] Explanation: No operation can be applied, we just shift the 0 to the end. Constraints: 2 <= nums.length <= 2000 0 <= nums[i] <= 1000

Explanation

Here's the solution to the problem:

• Apply Operations: Iterate through the array and perform the operations as described in the problem statement. If two adjacent elements are equal, multiply the first by 2 and set the second to 0.
• Shift Zeros: Create a new array to store the non-zero elements. Iterate through the modified array and add the non-zero elements to the new array.

• Pad with Zeros: After collecting all non-zero elements, pad the new array with the appropriate number of zeros to match the original array's length.

• Runtime Complexity: O(n), where n is the length of the input array.

• Storage Complexity: O(n), due to the creation of a new array to store the results. In place operations are possible but can be less readable.

Code

    def apply_operations(nums):
    n = len(nums)
    for i in range(n - 1):
        if nums[i] == nums[i + 1]:
            nums[i] *= 2
            nums[i + 1] = 0

    result = []
    zero_count = 0
    for num in nums:
        if num != 0:
            result.append(num)
        else:
            zero_count += 1

    result.extend([0] * zero_count)
    return result