What is the time complexity of Minimize Maximum of Array?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimize Maximum of Array?

This problem uses Array, Binary Search, Dynamic Programming, Greedy, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimize Maximum of Array?

Insight 1: The sum of the array remains constant after any number of operations. This is because we are only transferring values between elements, not adding or removing anything. Insight 2: We can frame the problem as finding the smallest value `x` such that we can redistribute values to the left to make all elements less than or equal to `x`.

Which companies ask LeetCode 2439. Minimize Maximum of Array in interviews?

This problem is frequently asked at Paytm.

The problem asks us to minimize the maximum value in an array `nums` by performing a series of operations. Each operation allows us to decrease an element `nums[i]` (where 1 0) by 1 and increase the preceding element `nums[i-1]` by 1. The goal is to find the smallest possible value that the maximum element in the array can be after applying any number of these operations.

Solution in python3. Solution in java. Solution in cpp. Solution in c

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Minimize Maximum of Array - LeetCode 2439 Solution

Minimize Maximum of Array - Complete Solution Guide

Minimize Maximum of Array is LeetCode problem 2439, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 beco

Detailed Explanation

The problem asks us to minimize the maximum value in an array `nums` by performing a series of operations. Each operation allows us to decrease an element `nums[i]` (where 1 <= i < n and `nums[i]` > 0) by 1 and increase the preceding element `nums[i-1]` by 1. The goal is to find the smallest possible value that the maximum element in the array can be after applying any number of these operations.

Solution Approach

The solution uses a greedy approach based on prefix sums. It iterates through the array, calculating the prefix sum at each index. For each index `i`, it computes the average value of the subarray from index 0 to `i` (inclusive). This average represents the minimum possible maximum value for that subarray if we were to redistribute the values optimally. We maintain a running maximum of these average values, which represents the overall minimum possible maximum value for the entire array.

Step-by-Step Algorithm

Step 1: Initialize `prefix_sum` to 0 and `result` to 0.
Step 2: Iterate through the `nums` array using a loop with index `i` and value `num`.
Step 3: Update `prefix_sum` by adding the current `num` to it: `prefix_sum += num`.
Step 4: Calculate the current average value. Since we need the ceiling of the average, we can add `i` to `prefix_sum` before dividing by `(i + 1)`: `current_avg_ceil = (prefix_sum + i) // (i + 1)` (Python) or `current_avg_ceil = (prefix_sum + i) / (i + 1)` (Java/C++)
Step 5: Update `result` to be the maximum of its current value and `current_avg_ceil`: `result = max(result, current_avg_ceil)`.
Step 6: After iterating through the entire array, return the final `result`.

Key Insights

Insight 1: The sum of the array remains constant after any number of operations. This is because we are only transferring values between elements, not adding or removing anything.
Insight 2: We can frame the problem as finding the smallest value `x` such that we can redistribute values to the left to make all elements less than or equal to `x`.
Insight 3: The core idea is to maintain a running prefix sum. For each index `i`, we want to find the minimum value `x` such that `prefix_sum[i] / (i+1)` (ceiling value) is less than or equal to `x`. This is because if we can make all elements up to index `i` have an average value of at most `x`, we can effectively redistribute the array.
Insight 4: The minimum maximum value will be the maximum of these average values for each prefix.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Dynamic Programming, Greedy, Prefix Sum.

Companies

Asked at: Paytm.

Minimize Maximum of Array | LeetCode 2439

Medium

Array Binary Search Dynamic Programming Greedy Prefix Sum

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minimizeArrayValue(int[] nums) {
        long prefix_sum = 0;
        long result = 0;
        for (int i = 0; i < nums.length; i++) {
            prefix_sum += nums[i];
            long current_avg_ceil = (prefix_sum + i) / (i + 1);
            result = Math.max(result, current_avg_ceil);
        }
        return (int) result;
    }
}

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Problem Info

Description

Problem 2439: Minimize Maximum of Array - Detailed explanation and solution approaches.

Topics

Array Binary Search Dynamic Programming Greedy Prefix Sum

Companies

Paytm

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2439

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize `prefix_sum` to 0 and `result` to 0.
2
Step 2: Iterate through the `nums` array using a loop with index `i` and value `num`.
3
Step 3: Update `prefix_sum` by adding the current `num` to it: `prefix_sum += num`.
4
Step 4: Calculate the current average value. Since we need the ceiling of the average, we can add `i` to `prefix_sum` before dividing by `(i + 1)`: `current_avg_ceil = (prefix_sum + i) // (i + 1)` (Python) or `current_avg_ceil = (prefix_sum + i) / (i + 1)` (Java/C++)
5
Step 5: Update `result` to be the maximum of its current value and `current_avg_ceil`: `result = max(result, current_avg_ceil)`.
6
Step 6: After iterating through the entire array, return the final `result`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because the algorithm iterates through the array only once. The space complexity is constant because the algorithm uses only a few variables to store the prefix sum and the result, regardless of the size of the input array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Binary Search

We can use binary search to find the minimum possible maximum value. The search space would be between the minimum and maximum element of the array. For each potential maximum value 'x', we would check if it's possible to make all elements less than or equal to 'x' by performing the operations. This involves iterating through the array and simulating the operations, keeping track of the 'debt' at each index. If the debt exceeds what can be satisfied, 'x' is too small.

Time: O(n log(sum(nums))), where n is the length of the array and sum(nums) is the sum of all elements in the array.Space: O(1)

Trade-offs: While Binary search might seem like a more intuitive approach at first, the given solution is more efficient due to O(n) time complexity. The binary search method requires a more complex implementation to check if a potential maximum is valid.

✨

Key Insights

1Insight 1: The sum of the array remains constant after any number of operations. This is because we are only transferring values between elements, not adding or removing anything.
2Insight 2: We can frame the problem as finding the smallest value `x` such that we can redistribute values to the left to make all elements less than or equal to `x`.
3Insight 3: The core idea is to maintain a running prefix sum. For each index `i`, we want to find the minimum value `x` such that `prefix_sum[i] / (i+1)` (ceiling value) is less than or equal to `x`. This is because if we can make all elements up to index `i` have an average value of at most `x`, we can effectively redistribute the array.
4Insight 4: The minimum maximum value will be the maximum of these average values for each prefix.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: Input array with only one element. The algorithm implicitly handles this case correctly since the loop starts from the beginning and the prefix sum and average are computed correctly for a single element.
Edge case 2: Array with all elements equal to 0. The algorithm handles this case as well. The prefix_sum will remain 0 and the current_avg_ceil will be 0 at each iteration resulting in 0 as the final result.
Edge case 3: Array with large numbers. The solution uses `long` (in Java/C++) or implicitly handles large integers (Python) for prefix_sum to prevent overflow.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using integer division instead of ceiling division can lead to incorrect results. Ensure that the average is rounded up correctly.

✗Common mistake 2: Not using long to store prefix_sum can lead to integer overflow issues when dealing with large input numbers, especially in languages like Java and C++.

✗Common mistake 3: Forgetting to account for the index `i` when calculating the ceiling of the average: (prefix_sum + i) / (i + 1).

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation. Imagine distributing tasks among workers, where each task has a certain workload. The problem can be mapped to minimizing the maximum workload any single worker has to deal with by re-allocating small portions of the workload between them.

🏦Real-world scenario 2: Load balancing in a server farm. Redistributing the load on each server to minimize the maximum load across all servers.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 8, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimize Maximum of Array"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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