LeetCode #3357 — HARD

Minimize the Maximum Adjacent Element Difference

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given an array of integers nums. Some values in nums are missing and are denoted by -1.

You must 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].
Patterns Used
🔍Modified Binary Search→🎯Greedy→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an array of integers nums. Some values in nums are missing and are denoted by -1. You must 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Greedy

Example 1

[1,2,-1,10,8]

Example 2

[-1,-1,-1]

Example 3

[-1,10,-1,8]

Related Problems

  • Minimum Absolute Sum Difference (minimum-absolute-sum-difference)
  • Minimize the Maximum Adjacent Element Difference (minimize-the-maximum-adjacent-element-difference)
Step 02

Core Insight

What unlocks the optimal approach

  • More than 2 occurrences of -1 can be ignored.
  • We can add the first positive number to the beginning and the last positive number to the end so that any consecutive of -1s are surrounded by positive numbers.
  • Suppose the answer is <code>d</code>, it can be proved that for the optimal case we'll replace -1s with values <code>0 < x <= y</code> and it's always optimal to select <code>x = min(a) + d</code>. So we only need to select <code>y</code>.
  • Binary search on <code>d</code>.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3357: Minimize the Maximum Adjacent Element Difference
class Solution {
  public int minDifference(int[] nums) {
    int maxPositiveGap = 0;
    int mn = 1_000_000_000;
    int mx = 0;

    for (int i = 1; i < nums.length; ++i) {
      if ((nums[i - 1] == -1) != (nums[i] == -1)) {
        final int positive = Math.max(nums[i - 1], nums[i]);
        mn = Math.min(mn, positive);
        mx = Math.max(mx, positive);
      } else {
        maxPositiveGap = Math.max(maxPositiveGap, Math.abs(nums[i - 1] - nums[i]));
      }
    }

    int l = maxPositiveGap;
    int r = (mx - mn + 1) / 2;

    while (l < r) {
      final int m = (l + r) / 2;
      if (check(nums, m, mn + m, mx - m))
        r = m;
      else
        l = m + 1;
    }

    return l;
  }

  // Returns true if it's possible have `m` as maximum absolute difference
  // between adjacent numbers, where -1s are replaced with `x` or `y`.
  private boolean check(int[] nums, int m, int x, int y) {
    int gapLength = 0;
    int prev = 0;

    for (final int num : nums) {
      if (num == -1) {
        ++gapLength;
        continue;
      }
      if (prev > 0 && gapLength > 0) {
        if (gapLength == 1 && !checkSingleGap(prev, num, m, x, y))
          return false;
        if (gapLength > 1 && !checkMultipleGaps(prev, num, m, x, y))
          return false;
      }
      prev = num;
      gapLength = 0;
    }

    // Check leading gaps
    if (nums[0] == -1) {
      final int num = findFirstNumber(nums, 0, 1);
      if (num != -1 && !checkBoundaryGaps(num, m, x, y))
        return false;
    }

    // Check trailing gaps
    if (nums[nums.length - 1] == -1) {
      final int num = findFirstNumber(nums, nums.length - 1, -1);
      if (num != -1 && !checkBoundaryGaps(num, m, x, y))
        return false;
    }

    return true;
  }

  // Returns true if it's possible to have at most `m` as the minimized maximum
  // difference for a sequence with a single -1 between two numbers.
  // e.g. [a, -1, b] can be filled with either x or y.
  private boolean checkSingleGap(int a, int b, int m, int x, int y) {
    final int gapWithX = Math.max(Math.abs(a - x), Math.abs(b - x)); // [a, x, b]
    final int gapWithY = Math.max(Math.abs(a - y), Math.abs(b - y)); // [a, y, b]
    return Math.min(gapWithX, gapWithY) <= m;
  }

  // Returns true if it's possible to have at most `m` as the minimized maximum
  // difference for a sequence with multiple -1s between two numbers.
  // e.g. [a, -1, -1, ..., -1, b] can be filled with x and y.
  private boolean checkMultipleGaps(int a, int b, int m, int x, int y) {
    final int ax = Math.abs(a - x);
    final int ay = Math.abs(a - y);
    final int bx = Math.abs(b - x);
    final int by = Math.abs(b - y);
    final int xy = Math.abs(x - y);
    final int gapAllX = Math.max(ax, bx);               // [a, x, x, ..., x, b]
    final int gapAllY = Math.max(ay, by);               // [a, y, y, ..., y, b]
    final int gapXToY = Math.max(Math.max(ax, xy), by); // [a, x, ..., y, b]
    final int gapYToX = Math.max(Math.max(ay, xy), bx); // [a, y, ..., x, b]
    return Math.min(Math.min(gapAllX, gapAllY), Math.min(gapXToY, gapYToX)) <= m;
  }

  // Returns true if it's possible to have at most `m` as the minimized maximum
  // difference for a boundary sequence starting or ending with -1s.
  // e.g. [a, -1, -1, ...] or [..., -1, -1, a].
  private boolean checkBoundaryGaps(int a, int m, int x, int y) {
    final int gapAllX = Math.abs(a - x); // [x, x, ..., x, a]
    final int gapAllY = Math.abs(a - y); // [y, y, ..., y, a]
    return Math.min(gapAllX, gapAllY) <= m;
  }

  // Returns the first positive number starting from the given index or -1
  // if not found.
  private int findFirstNumber(int[] nums, int start, int step) {
    int i = start;
    while (i >= 0 && i < nums.length && nums[i] == -1)
      i += step;
    return i >= 0 && i < nums.length ? nums[i] : -1;
  }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(log n)
Space
O(1)

Approach Breakdown

LINEAR SCAN
O(n) time
O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH
O(log n) time
O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.

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