What is the time complexity of Minimize Deviation in Array?

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

What data structures are used to solve Minimize Deviation in Array?

This problem uses Array, Greedy, Heap (Priority Queue), Ordered Set. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimize Deviation in Array?

Insight 1: To minimize the deviation, we want to bring the maximum and minimum elements of the array as close as possible. Insight 2: Each odd number can only be multiplied by 2 once. Each even number can be divided by 2 multiple times until it becomes odd or 1.

Which companies ask LeetCode 1675. Minimize Deviation in Array in interviews?

This problem is frequently asked at Samsung.

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Minimize Deviation in Array - LeetCode 1675 Solution

Minimize Deviation in Array - Complete Solution Guide

Minimize Deviation in Array is LeetCode problem 1675, a Hard 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 an array nums of n positive integers. You can perform two types of operations on any element of the array any number of times: If the element is even , divide it by 2 . For example, if the array is [1,2,3,4] , then you can do this operation on the last element, and the array will be [1,2,3, 2 ]. If the element is odd , multiply it by 2 . For example, if the array is [1,2,3,4] , then you can do this operation on the first element, and the array will be [ 2 ,2,3,4]. The deviation of

Detailed Explanation

The problem asks us to minimize the 'deviation' of an array of positive integers. Deviation is defined as the maximum difference between any two elements in the array. We are allowed to perform two types of operations any number of times: divide an even number by 2, or multiply an odd number by 2. The goal is to find the smallest possible deviation we can achieve after applying these operations.

Solution Approach

The solution uses a priority queue (max heap) to efficiently track the current maximum element and allows us to repeatedly reduce the largest element by dividing it by 2 until it becomes odd or reaches its minimum possible value (which is either the original value if it started even, or double the value if it started odd). We also maintain a running minimum value, as dividing the maximum may result in a new, smaller minimum. The difference between the current maximum and current minimum represents the deviation at any given moment, and we want to minimize this.

Step-by-Step Algorithm

Step 1: Transform all odd numbers in the array by multiplying them by 2. This makes them even and ensures we only perform division operations going forward.
Step 2: Insert all numbers into a max priority queue. Use a negative representation of the numbers so a min-heap is used as a max-heap.
Step 3: Find the initial minimum value among all the numbers after transformation.
Step 4: Initialize the minimum deviation as the difference between the current maximum (top of the heap) and the minimum value.
Step 5: While the current maximum element (top of the heap) is even:
a. Remove the current maximum from the heap.
b. Divide the current maximum by 2.
c. Insert the new value back into the heap.
d. Update the minimum value if the new value is smaller than the current minimum.
e. Update the minimum deviation if the current deviation is smaller than the previous minimum deviation.
Step 6: Return the minimum deviation found.

Key Insights

Insight 1: To minimize the deviation, we want to bring the maximum and minimum elements of the array as close as possible.
Insight 2: Each odd number can only be multiplied by 2 once. Each even number can be divided by 2 multiple times until it becomes odd or 1.
Insight 3: We can maximize each number initially (multiply odd numbers by 2). This ensures that we are only ever decreasing numbers (dividing by 2). The minimum element needs to be tracked separately.

Complexity Analysis

Time Complexity: O(n log n log M)

Space Complexity: O(n)

Topics

This problem involves: Array, Greedy, Heap (Priority Queue), Ordered Set.

Companies

Asked at: Samsung.

Minimize Deviation in Array | LeetCode 1675

Hard

Array Greedy Heap (Priority Queue)Ordered Set

Time: O(n log n log M)Space: O(n)

Solution Code

Available in:

import java.util.PriorityQueue;

class Solution {
    public int minimumDeviation(int[] nums) {
        PriorityQueue<Integer> pq = new PriorityQueue<>();
        int min_val = Integer.MAX_VALUE;
        for (int n : nums) {
            if (n % 2 == 1) {
                pq.offer(-n * 2);
                min_val = Math.min(min_val, n * 2);
            } else {
                pq.offer(-n);
                min_val = Math.min(min_val, n);
            }
        }

        int min_deviation = -pq.peek() - min_val;

        while (pq.peek() % 2 == 0) {
            int current_max = -pq.poll();
            int new_val = current_max / 2;
            pq.offer(-new_val);

            min_val = Math.min(min_val, new_val);
            min_deviation = Math.min(min_deviation, -pq.peek() - min_val);
        }

        return min_deviation;
    }
}

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Problem Info

Description

Problem 1675: Minimize Deviation in Array - Detailed explanation and solution approaches.

Topics

Array Greedy Heap (Priority Queue)Ordered Set

Companies

Samsung

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1675

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

11 steps

1
Step 1: Transform all odd numbers in the array by multiplying them by 2. This makes them even and ensures we only perform division operations going forward.
2
Step 2: Insert all numbers into a max priority queue. Use a negative representation of the numbers so a min-heap is used as a max-heap.
3
Step 3: Find the initial minimum value among all the numbers after transformation.
4
Step 4: Initialize the minimum deviation as the difference between the current maximum (top of the heap) and the minimum value.
5
Step 5: While the current maximum element (top of the heap) is even:
6
a. Remove the current maximum from the heap.
7
b. Divide the current maximum by 2.
8
c. Insert the new value back into the heap.
9
d. Update the minimum value if the new value is smaller than the current minimum.
10
e. Update the minimum deviation if the current deviation is smaller than the previous minimum deviation.
11
Step 6: Return the minimum deviation found.

💡

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 log n log M)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the heap operations and the potential number of divisions. The space complexity is due to storing the numbers in a heap.

📖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

Using a Sorted Set

Instead of a priority queue, we can use a sorted set (e.g., TreeSet in Java or set in C++). The advantage is that we can directly access both the maximum and minimum elements in O(log n) time. The core logic remains the same: repeatedly divide the maximum element by 2 and update the set, while keeping track of the global minimum.

Time: O(n log n log M), similar to the priority queue approach.Space: O(n), same as the priority queue approach.

Trade-offs: Sorted set might offer slightly better performance in some implementations due to optimized search and update operations. However, the overall complexity remains the same.

✨

Key Insights

1Insight 1: To minimize the deviation, we want to bring the maximum and minimum elements of the array as close as possible.
2Insight 2: Each odd number can only be multiplied by 2 once. Each even number can be divided by 2 multiple times until it becomes odd or 1.
3Insight 3: We can maximize each number initially (multiply odd numbers by 2). This ensures that we are only ever decreasing numbers (dividing by 2). The minimum element needs to be tracked separately.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: Input array contains only even numbers. The algorithm correctly handles this by repeatedly dividing the maximum until it potentially becomes odd or reaches its smallest possible value.
Edge case 2: Input array contains only odd numbers. The algorithm correctly handles this by multiplying all odd numbers by 2 initially and then proceeding as usual.
Edge case 3: Input array contains only the number 1. The algorithm correctly multiplies it by 2 and the maximum and minimum are the same, resulting in deviation of 0.
Edge case 4: Array containing a single element. This case is handled correctly as the max and min are the element, and the deviation will be zero.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not multiplying odd numbers by 2 initially. This leads to incorrect deviation calculations and may prevent the algorithm from finding the optimal solution.

✗Common mistake 2: Not tracking the minimum value correctly. After each division, the minimum value might change. Failing to update the minimum value can lead to a larger deviation.

✗Common mistake 3: Integer overflow. When multiplying odd numbers by two, it is essential to be mindful of the constraints and potential overflows. While the problem constraints make this less of a concern, it is something to be aware of for similar problems.

✗Common mistake 4: Not considering edge cases. Consider single element arrays or arrays containing just even or just odd numbers.

✅

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 allocating resources with varying initial quantities. Operations can either combine resources (multiply) or split resources (divide). The goal is to minimize the difference between the largest and smallest resource allocations.

🏦Real-world scenario 2: Data compression. The algorithm could be used to minimize the range of values in a dataset by applying transformations. Minimizing this range can improve compression ratios or reduce storage requirements.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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: July 26, 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 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.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 Deviation in Array"?

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

🤖 Need help with this problem?