LeetCode #2439 — MEDIUM

Minimize Maximum of Array

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
The Problem

Problem Statement

You are given a 0-indexed array nums comprising of n non-negative integers.

In one operation, you must:

  • Choose an integer i such that 1 <= i < n and nums[i] > 0.
  • Decrease nums[i] by 1.
  • Increase nums[i - 1] by 1.

Return the minimum possible value of the maximum integer of nums after performing any number of operations.

Example 1:

Input: nums = [3,7,1,6]
Output: 5
Explanation:
One set of optimal operations is as follows:
1. Choose i = 1, and nums becomes [4,6,1,6].
2. Choose i = 3, and nums becomes [4,6,2,5].
3. Choose i = 1, and nums becomes [5,5,2,5].
The maximum integer of nums is 5. It can be shown that the maximum number cannot be less than 5.
Therefore, we return 5.

Example 2:

Input: nums = [10,1]
Output: 10
Explanation:
It is optimal to leave nums as is, and since 10 is the maximum value, we return 10.

Constraints:

  • n == nums.length
  • 2 <= n <= 105
  • 0 <= nums[i] <= 109
Patterns Used
🔍Modified Binary Search→📊Dynamic Programming→🎯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 a 0-indexed array nums comprising of n non-negative integers. In one operation, you must: Choose an integer i such that 1 <= i < n and nums[i] > 0. Decrease nums[i] by 1. Increase nums[i - 1] by 1. Return the minimum possible value of the maximum integer of nums after performing any number of operations.

Baseline thinking

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

Pattern signal: Array · Binary Search · Dynamic Programming · Greedy

Example 1

[3,7,1,6]

Example 2

[10,1]

Related Problems

  • Maximum Candies Allocated to K Children (maximum-candies-allocated-to-k-children)
  • Minimum Speed to Arrive on Time (minimum-speed-to-arrive-on-time)
  • Minimum Time to Complete Trips (minimum-time-to-complete-trips)
Step 02

Core Insight

What unlocks the optimal approach

  • Try a binary search approach.
  • Perform a binary search over the minimum value that can be achieved for the maximum number of the array.
  • In each binary search iteration, iterate through the array backwards, greedily decreasing the current element until it is within the limit.
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
Upper-end input sizes
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 #2439: Minimize Maximum of Array
class Solution {
    private int[] nums;

    public int minimizeArrayValue(int[] nums) {
        this.nums = nums;
        int left = 0, right = max(nums);
        while (left < right) {
            int mid = (left + right) >> 1;
            if (check(mid)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

    private boolean check(int mx) {
        long d = 0;
        for (int i = nums.length - 1; i > 0; --i) {
            d = Math.max(0, d + nums[i] - mx);
        }
        return nums[0] + d <= mx;
    }

    private int max(int[] nums) {
        int v = nums[0];
        for (int x : nums) {
            v = Math.max(v, x);
        }
        return v;
    }
}
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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

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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