Introduction

In this blog post, we will explore how to solve the LeetCode problem "3357" 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 an array of integers nums. Some values in nums are missing and are denoted by -1. You can choose a pair of positive integers (x, y) exactly once and replace each missing element with either x or y. You need to minimize the maximum absolute difference between adjacent elements of nums after replacements. Return the minimum possible difference. Example 1: Input: nums = [1,2,-1,10,8] Output: 4 Explanation: By choosing the pair as (6, 7), nums can be changed to [1, 2, 6, 10, 8]. The absolute differences between adjacent elements are: |1 - 2| == 1 |2 - 6| == 4 |6 - 10| == 4 |10 - 8| == 2 Example 2: Input: nums = [-1,-1,-1] Output: 0 Explanation: By choosing the pair as (4, 4), nums can be changed to [4, 4, 4]. Example 3: Input: nums = [-1,10,-1,8] Output: 1 Explanation: By choosing the pair as (11, 9), nums can be changed to [11, 10, 9, 8]. Constraints: 2 <= nums.length <= 105 nums[i] is either -1 or in the range [1, 109].

Explanation

Here's the approach, followed by the code:

• Binary Search: The core idea is to use binary search to find the minimum possible maximum absolute difference. We search within the range of possible differences (0 to a large value, e.g., 10^9).
• Check Feasibility: For each potential maximum difference mid during binary search, we check if it's possible to replace the -1 values such that the maximum absolute difference between adjacent elements never exceeds mid. This check is done by iterating through the array and strategically choosing either x or y for each -1 to meet the mid constraint.

• Determine x and y: To check if the mid is feasible we will find the possible ranges [low, high] for x and y and then check if there exist x and y within that range which makes this possible.

• Time Complexity: O(n log m), where n is the length of the input array and m is the range of possible absolute differences. Space Complexity: O(1).

def solve():
    def check(nums, mid, x, y):
        prev = None
        for i in range(len(nums)):
            if nums[i] != -1:
                if prev is not None and abs(prev - nums[i]) > mid:
                    return False
                prev = nums[i]

        return True

    def check_feasibility(nums, mid):
        min_val = float('inf')
        max_val = float('-inf')

        for i in range(len(nums) - 1):
            if nums[i] != -1 and nums[i+1] != -1:
                continue

            if nums[i] == -1 and nums[i+1] == -1:
                continue

            if nums[i] == -1:
                if min_val > nums[i+1] - mid:
                    min_val = nums[i+1] - mid
                if max_val < nums[i+1] + mid:
                    max_val = nums[i+1] + mid

            elif nums[i+1] == -1:
                if min_val > nums[i] - mid:
                    min_val = nums[i] - mid
                if max_val < nums[i] + mid:
                    max_val = nums[i] + mid


        if min_val == float('inf') and max_val == float('-inf'):
            return True


        return min_val <= max_val

    nums = list(map(int, input().split()))

    left = 0
    right = 10**9
    ans = right

    while left <= right:
        mid = (left + right) // 2
        if check_feasibility(nums, mid):
            ans = mid
            right = mid - 1
        else:
            left = mid + 1

    print(ans)

# Example Usage (you'd normally run this from a script, not an online judge):
# nums = [1, 2, -1, 10, 8]
# solve()

# nums = [-1, -1, -1]
# solve()

# nums = [-1, 10, -1, 8]
# solve()

def solve_wrapper():
  nums = list(map(int, input().split()))
  def check_feasibility(nums, mid):
        gaps = []
        for i in range(len(nums) - 1):
            if nums[i] == -1 or nums[i+1] == -1:
                gaps.append((i, i+1))

        if not gaps:
            max_diff = 0
            for i in range(len(nums) - 1):
                max_diff = max(max_diff, abs(nums[i] - nums[i+1]))
            return max_diff <= mid

        low = float('-inf')
        high = float('inf')

        for i in range(len(nums) - 1):
            if nums[i] != -1 and nums[i+1] != -1:
                if abs(nums[i] - nums[i+1]) > mid:
                    return False
            elif nums[i] == -1 and nums[i+1] != -1:
                low = max(low, nums[i+1] - mid)
                high = min(high, nums[i+1] + mid)
            elif nums[i] != -1 and nums[i+1] == -1:
                low = max(low, nums[i] - mid)
                high = min(high, nums[i] + mid)

        if low > high:
            return False

        return True

  left = 0
  right = int(2e9)
  ans = right
  while left <= right:
      mid = (left + right) // 2
      if check_feasibility(nums, mid):
          ans = mid
          right = mid - 1
      else:
          left = mid + 1
  print(ans)


    # Code
    ```python
    def solve_wrapper():  nums = list(map(int, input().split()))
  def check_feasibility(nums, mid):
        gaps = []
        for i in range(len(nums) - 1):
            if nums[i] == -1 or nums[i+1] == -1:
                gaps.append((i, i+1))

        if not gaps:
            max_diff = 0
            for i in range(len(nums) - 1):
                max_diff = max(max_diff, abs(nums[i] - nums[i+1]))
            return max_diff <= mid

        low = float('-inf')
        high = float('inf')

        for i in range(len(nums) - 1):
            if nums[i] != -1 and nums[i+1] != -1:
                if abs(nums[i] - nums[i+1]) > mid:
                    return False
            elif nums[i] == -1 and nums[i+1] != -1:
                low = max(low, nums[i+1] - mid)
                high = min(high, nums[i+1] + mid)
            elif nums[i] != -1 and nums[i+1] == -1:
                low = max(low, nums[i] - mid)
                high = min(high, nums[i] + mid)

        if low > high:
            return False

        return True

  left = 0
  right = int(2e9)
  ans = right
  while left <= right:
      mid = (left + right) // 2
      if check_feasibility(nums, mid):
          ans = mid
          right = mid - 1
      else:
          left = mid + 1
  print(ans)