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Minimize the Maximum Adjacent Element Difference

Number: 3658

Difficulty: Hard

Paid? No

Companies: N/A

Problem Description

You are given an integer array nums where some entries are missing and represented by –1. You are allowed to choose a pair of positive integers (x, y) exactly once and then replace every missing element (–1) with either x or y – you may choose independently for each missing position. After performing the replacements the score of the array is defined as the maximum absolute difference between any two adjacent elements. The goal is to minimize that score and return the minimum possible maximum difference.

For example, if nums = [1,2,-1,10,8] one optimal solution is to choose (6,7) and replace –1 by 6 so that the resulting array is [1,2,6,10,8] whose adjacent differences are [1,4,4,2] and the maximum is 4.

Key Insights

  • Only adjacent differences matter. Differences between two non–missing (fixed) numbers are set in stone.
  • For each contiguous group of missing elements, you may assign a single value (either x or y) to all positions in that group. (Even if the replacement pair allows different numbers globally, within one group it is always best to use a uniform number to avoid additional missing–to–missing differences.)
  • If a missing group is flanked by two fixed values L and R then, when filled with one number v, the cost contributed by that group is max(|L-v|, |R-v|). The optimal choice is to pick v in between L and R so that the optimal cost is ceil(|L-R|/2).
  • For groups with only one fixed neighbor (i.e. at the beginning or end), you can “match” the fixed neighbor exactly.
  • Therefore the answer is the maximum of: • the maximum difference among adjacent fixed numbers (which you cannot change), and • the worst cost (over all groups flanked by two fixed numbers) computed as ceil(|L-R|/2).

Space and Time Complexity

Time Complexity: O(n) – We iterate through the array a constant number of times. Space Complexity: O(1) – Only a constant amount of extra space is used.

Solution

The idea is to scan the array to gather two key pieces of information. First, compute max_fixed_diff as the maximum absolute difference between adjacent numbers that are both not –1. These differences are immutable. Second, for each contiguous group of –1’s that is flanked on both sides by fixed numbers, compute the cost needed if that missing block is replaced uniformly with a number optimally chosen between the two fixed values. This cost equals ⌈|L - R| / 2⌉ (which can be computed as (|L-R|+1)//2 in integer arithmetic). For groups that lie at an edge (only one neighbor), the cost can be made 0 by matching the fixed neighbor. Finally the answer is the maximum of the immutable max_fixed_diff and the largest cost computed among all two–boundary groups. This works because although you have two numbers available for replacement (x and y), different missing segments are separated by fixed numbers so the decision can be made independently in each segment.

Data structures: We only use integer variables to store the interim results. Algorithmic approach: A single iteration (or two) over the array is sufficient.

Code Solutions

# Python solution with detailed comments
def minimizeMaxDifference(nums):
    n = len(nums)
    max_fixed_diff = 0    # maximum difference between adjacent non -1 numbers
    ans_missing_segment = 0  # required cost for segments with two fixed ends

    # First pass: compute max_fixed_diff for adjacent fixed numbers.
    for i in range(1, n):
        if nums[i] != -1 and nums[i-1] != -1:
            diff = abs(nums[i] - nums[i-1])
            if diff > max_fixed_diff:
                max_fixed_diff = diff

    i = 0
    while i < n:
        # If current element is not missing, just proceed.
        if nums[i] != -1:
            i += 1
            continue
        # Found a contiguous group of missing elements.
        start = i
        # move pointer until we exit the group of -1's.
        while i < n and nums[i] == -1:
            i += 1
        end = i - 1  # group from index start to end
        # Determine fixed border values if they exist.
        left_value = nums[start - 1] if start - 1 >= 0 else None
        right_value = nums[i] if i < n else None
        # Only if both borders exist we require a compromise value.
        if left_value is not None and right_value is not None:
            # The optimal cost for assigning one uniform value is ceil(|L - R|/2)
            segment_cost = (abs(left_value - right_value) + 1) // 2
            ans_missing_segment = max(ans_missing_segment, segment_cost)
        # If only one border exists, we can always match that border exactly so cost is 0.
    
    # The answer is the maximum of the immutable differences and the best needed cost
    return max(max_fixed_diff, ans_missing_segment)

# Example usage:
print(minimizeMaxDifference([1,2,-1,10,8]))  # Expected output: 4
print(minimizeMaxDifference([-1,-1,-1]))       # Expected output: 0
print(minimizeMaxDifference([-1,10,-1,8]))       # Expected output: 1

# For interactive judges, you would typically read input and output accordingly.
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