Introduction

In this blog post, we will explore how to solve the LeetCode problem "645" 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 have a set of integers s, which originally contains all the numbers from 1 to n. Unfortunately, due to some error, one of the numbers in s got duplicated to another number in the set, which results in repetition of one number and loss of another number. You are given an integer array nums representing the data status of this set after the error. Find the number that occurs twice and the number that is missing and return them in the form of an array. Example 1: Input: nums = [1,2,2,4] Output: [2,3] Example 2: Input: nums = [1,1] Output: [1,2] Constraints: 2 <= nums.length <= 104 1 <= nums[i] <= 104

Explanation

Here's the breakdown of the solution:

• Utilize the array itself as a hash table: Iterate through the array. For each number num, negate the element at index abs(num) - 1. If the element at that index is already negative, it means num is the duplicate.
• Find the missing number: After the first pass, iterate through the array again. The index of the positive number corresponds to the missing number.

• Handle edge case: If no positive number is found after the second pass, the missing number is n

• Time and Space Complexity: O(n) time complexity and O(1) space complexity (excluding the output array).

Code

    def findErrorNums(nums):
    """
    Finds the duplicate and missing numbers in an array.

    Args:
        nums: A list of integers representing the array with one duplicate and one missing number.

    Returns:
        A list containing the duplicate and missing numbers.
    """
    n = len(nums)
    duplicate = -1
    missing = -1

    # Find the duplicate number
    for i in range(n):
        index = abs(nums[i]) - 1
        if nums[index] < 0:
            duplicate = abs(nums[i])
        else:
            nums[index] *= -1

    # Find the missing number
    for i in range(n):
        if nums[i] > 0:
            missing = i + 1
            break

    # If no positive number is found, the missing number is n
    if missing == -1:
        missing = n

    return [duplicate, missing]