Introduction

In this blog post, we will explore how to solve the LeetCode problem "1920" 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

Given a zero-based permutation nums (0-indexed), build an array ans of the same length where ans[i] = nums[nums[i]] for each 0 <= i < nums.length and return it. A zero-based permutation nums is an array of distinct integers from 0 to nums.length - 1 (inclusive). Example 1: Input: nums = [0,2,1,5,3,4] Output: [0,1,2,4,5,3] Explanation: The array ans is built as follows: ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]] = [nums[0], nums[2], nums[1], nums[5], nums[3], nums[4]] = [0,1,2,4,5,3] Example 2: Input: nums = [5,0,1,2,3,4] Output: [4,5,0,1,2,3] Explanation: The array ans is built as follows: ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]] = [nums[5], nums[0], nums[1], nums[2], nums[3], nums[4]] = [4,5,0,1,2,3] Constraints: 1 <= nums.length <= 1000 0 <= nums[i] < nums.length The elements in nums are distinct. Follow-up: Can you solve it without using an extra space (i.e., O(1) memory)?

Explanation

Here's the breakdown of the solution:

• In-place modification: The core idea is to modify the original array nums to store the desired result without using extra space. We achieve this by encoding the result in the existing array's elements.
• Encoding and decoding: We encode the desired value nums[nums[i]] into nums[i] by using modulo and division operations. Specifically, we store nums[nums[i]] by adding nums[nums[i]] % n * n to nums[i]. Later, we can extract the encoded value by dividing nums[i] by n.

• Iterate and decode: Finally, we iterate through the modified array and decode the values to get the result in the correct format.

• Runtime Complexity: O(n), Storage Complexity: O(1)

Code

    def buildArray(nums):
    n = len(nums)

    # Encoding phase: Store nums[nums[i]] in nums[i]
    for i in range(n):
        nums[i] += (nums[nums[i]] % n) * n

    # Decoding phase: Extract the stored values
    for i in range(n):
        nums[i] //= n

    return nums