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Minimize the Maximum Adjacent Element Difference

Difficulty: Hard

Problem Description

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.

Key Insights

  • The problem involves replacing missing values (-1) with two chosen positive integers (x, y) to minimize the maximum difference between adjacent elements.
  • The main challenge is determining the optimal values for (x, y) such that the maximum difference is minimized.
  • The problem can be solved using binary search to efficiently find the minimum possible maximum difference.

Space and Time Complexity

Time Complexity: O(n log m) where n is the number of elements in the array and m is the maximum possible difference. Space Complexity: O(1) as we only use a constant amount of additional space.

Solution

To solve this problem, we can use a binary search approach. The idea is to determine the minimum possible maximum difference by checking if a certain maximum difference is achievable with valid choices of (x, y).

  1. Identify Bounds: First, we need to determine the minimum and maximum values in the array, ignoring the missing values (-1).
  2. Binary Search Setup: We will perform binary search on the possible maximum differences, starting from 0 to the difference between the maximum and minimum known values.
  3. Feasibility Check: For each mid-point value in the binary search, we check if it's possible to replace -1s with values such that no adjacent differences exceed this mid-point.
  4. Adjust Search Range: Depending on whether the current mid-point is feasible, we adjust our binary search bounds accordingly.

This approach ensures we efficiently explore the solution space to find the optimal pair (x, y).

Code Solutions

def minimizeMaxDifference(nums):
    def canAchieve(maxDiff):
        last = -1
        for num in nums:
            if num != -1:
                if last != -1 and abs(num - last) > maxDiff:
                    return False
                last = num
            else:
                # We have two choices for -1: choose x or y
                # We should ensure that the gaps can be filled
                if last != -1:
                    # Fill with a number within maxDiff range
                    if last + maxDiff >= 1:
                        last = last + maxDiff
                    else:
                        last = 1  # choosing a minimum positive integer
                else:
                    last = 1  # first element can be any positive integer
        return True
    
    low, high = 0, 10**9
    while low < high:
        mid = (low + high) // 2
        if canAchieve(mid):
            high = mid
        else:
            low = mid + 1
    return low
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