What is the time complexity of Minimum Moves to Equal Array Elements II?

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

What data structures are used to solve Minimum Moves to Equal Array Elements II?

This problem uses Array, Math, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Moves to Equal Array Elements II?

Insight 1: The optimal target value to minimize the total moves is the median of the array. Moving all elements towards the median minimizes the total distance. Insight 2: Sorting the array is essential to efficiently find the median. For an array of size n, the median is located at index n/2 after sorting (for both even and odd length arrays; with even length it will give one of the two medians, either is ok).

Which companies ask LeetCode 462. Minimum Moves to Equal Array Elements II in interviews?

This problem is frequently asked at Myntra.

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Minimum Moves to Equal Array Elements II - LeetCode 462 Solution

Minimum Moves to Equal Array Elements II - Complete Solution Guide

Minimum Moves to Equal Array Elements II is LeetCode problem 462, 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

Given an integer array nums of size n , return the minimum number of moves required to make all array elements equal . In one move, you can increment or decrement an element of the array by 1 . Test cases are designed so that the answer will fit in a 32-bit integer. Example 1: Input: nums = [1,2,3] Output: 2 Explanation: Only two moves are needed (remember each move increments or decrements one element): [ 1 ,2,3] => [2,2, 3 ] => [2,2,2] Example 2: Input: nums = [1,10,2,9] Output: 16 Constraints

Detailed Explanation

The problem asks us to find the minimum number of moves needed to make all elements in an integer array equal. A single move consists of incrementing or decrementing any element by 1. The goal is to minimize the sum of differences between each element and a target value, which will be the final value of all elements after the moves.

Solution Approach

The solution calculates the minimum moves by first sorting the input array. Then, it identifies the median element in the sorted array. The number of moves required for each element is the absolute difference between the element and the median. The total number of moves is the sum of these absolute differences.

Step-by-Step Algorithm

Step 1: Sort the input array `nums` in ascending order using an efficient sorting algorithm (e.g., quicksort, mergesort) available in the respective language's standard library.
Step 2: Calculate the index of the median. This is `len(nums) // 2` in Python and `nums.length / 2` in Java and C++.
Step 3: Retrieve the median element using the calculated index: `median = nums[index]`.
Step 4: Initialize a variable `moves` to 0. This variable will accumulate the total number of moves.
Step 5: Iterate through the array `nums`. For each element `num`, calculate the absolute difference between `num` and the `median`: `abs(num - median)`. Add this difference to the `moves` variable.
Step 6: After iterating through all elements, the `moves` variable contains the minimum number of moves required to make all elements equal. Return the `moves`.

Key Insights

Insight 1: The optimal target value to minimize the total moves is the median of the array. Moving all elements towards the median minimizes the total distance.
Insight 2: Sorting the array is essential to efficiently find the median. For an array of size n, the median is located at index n/2 after sorting (for both even and odd length arrays; with even length it will give one of the two medians, either is ok).
Insight 3: The problem guarantees that the result will fit in a 32-bit integer, which means we don't have to worry about integer overflow during the summation of moves.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Sorting.

Companies

Asked at: Myntra.

Minimum Moves to Equal Array Elements II | LeetCode 462

Medium

Array Math Sorting

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int minMoves2(int[] nums) {
        Arrays.sort(nums);
        int median = nums[nums.length / 2];
        int moves = 0;
        for (int num : nums) {
            moves += Math.abs(num - median);
        }
        return moves;
    }
}

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Problem Info

Description

Problem 462: Minimum Moves to Equal Array Elements II - Detailed explanation and solution approaches.

Topics

Array Math Sorting

Companies

Myntra

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:462

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: Sort the input array `nums` in ascending order using an efficient sorting algorithm (e.g., quicksort, mergesort) available in the respective language's standard library.
2
Step 2: Calculate the index of the median. This is `len(nums) // 2` in Python and `nums.length / 2` in Java and C++.
3
Step 3: Retrieve the median element using the calculated index: `median = nums[index]`.
4
Step 4: Initialize a variable `moves` to 0. This variable will accumulate the total number of moves.
5
Step 5: Iterate through the array `nums`. For each element `num`, calculate the absolute difference between `num` and the `median`: `abs(num - median)`. Add this difference to the `moves` variable.
6
Step 6: After iterating through all elements, the `moves` variable contains the minimum number of moves required to make all elements equal. Return the `moves`.

💡

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-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the sorting step, which typically takes O(n log n) time using efficient sorting algorithms like quicksort or mergesort. The remaining steps, such as finding the median and calculating the sum of absolute differences, take O(n) time. Therefore, the overall time complexity is O(n log n). The space complexity is O(1) because the algorithm operates in place (or with a small constant amount of extra space), without creating any data structures that scale linearly with the input size, assuming that the sorting algorithm used is in-place or requires only logarithmic space.

📖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

Quickselect Algorithm

Instead of sorting the entire array, we can use the Quickselect algorithm to find the median in O(n) average time complexity. This would improve the overall time complexity.

Time: O(n) average, O(n^2) worst-caseSpace: O(1)

Trade-offs: Quickselect can be faster on average, but it has a worst-case time complexity of O(n^2). The standard `sort` method typically guarantees O(n log n) complexity, making it more predictable in performance. For practical purposes, the O(n log n) sorting may be preferable due to its guaranteed upper bound. Choosing between sorting and quickselect depends on the specific performance requirements and risk tolerance for the worst-case scenario.

✨

Key Insights

1Insight 1: The optimal target value to minimize the total moves is the median of the array. Moving all elements towards the median minimizes the total distance.
2Insight 2: Sorting the array is essential to efficiently find the median. For an array of size n, the median is located at index n/2 after sorting (for both even and odd length arrays; with even length it will give one of the two medians, either is ok).
3Insight 3: The problem guarantees that the result will fit in a 32-bit integer, which means we don't have to worry about integer overflow during the summation of moves.

📚

Key Terms Explained

For Students

Array

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty Array: Although the constraint specifies that the array length is at least 1, handling an empty array gracefully is a good practice. We could add a condition at the beginning to return 0 if `nums` is empty to avoid errors.
Edge case 2: Array with one element: If the array contains only one element, no moves are required. The algorithm correctly handles this case because the median is the only element, and the absolute difference between the element and itself is 0.
Edge case 3: Array with all same elements: If all elements in the array are the same, the median will be equal to all elements, and the total number of moves will be 0, which is the correct answer.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the array: Forgetting to sort the array before finding the median will result in an incorrect median value and, therefore, an incorrect number of moves.

✗Common mistake 2: Incorrectly calculating the median index: Off-by-one errors when calculating the median index can lead to incorrect results. Make sure to use integer division to find the correct index, especially in languages like Python.

✗Common mistake 3: Using the mean (average) instead of the median: The average is not always the best target value to minimize the total moves. The median is the correct target value for this type of problem because it minimizes the sum of absolute deviations.

✗Common mistake 4: Integer Overflow: Although the problem states that the answer fits in a 32-bit integer, not handling large numbers carefully in intermediate calculations (specifically when summing the absolute differences) *could* lead to integer overflows in other, similar problems, especially in languages like Java or C++. It's a good practice to be mindful of this even when not explicitly required by the problem constraints. Using `long` to store the sum can help avoid this.

✅

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: Data balancing in machine learning: In some machine learning scenarios, we might want to balance the distribution of data points by moving data points closer to a central value, where each move has a cost. This problem models such scenarios.

🏦Real-world scenario 2: Resource allocation: Imagine distributing resources (e.g., bandwidth, computing power) among users such that the total cost of reallocation is minimized. Each user's resource consumption can be represented by an element in the array, and the cost of reallocation is the absolute difference between the allocated resource and the target resource level.

💡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 11, 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

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

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Moves to Equal Array Elements II"?

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 math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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