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

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

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

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

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

Insight 1: Incrementing `n - 1` elements is equivalent to decrementing only one element. This is the core insight that simplifies the problem. Insight 2: The target value for all elements to converge to is the minimum value in the original array. Decrementing all other elements until they equal the minimum value is the optimal strategy.

Which companies ask LeetCode 453. Minimum Moves to Equal Array Elements in interviews?

This problem is frequently asked at Coursera, Indeed.

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Minimum Moves to Equal Array Elements - LeetCode 453 Solution

Q: How to solve Minimum Moves to Equal Array Elements?

The provided solution leverages the insight that incrementing `n - 1` elements is equivalent to decrementing one element. Therefore, instead of thinking about incrementing multiple elements, the problem can be reframed as finding the minimum number of decrements needed to make all elements equal to the minimum value in the array. The solution calculates the sum of differences between each element and the minimum element. This sum represents the total number of moves required.

Minimum Moves to Equal Array Elements - Complete Solution Guide

Minimum Moves to Equal Array Elements is LeetCode problem 453, 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 n - 1 elements of the array by 1 . Example 1: Input: nums = [1,2,3] Output: 3 Explanation: Only three moves are needed (remember each move increments two elements): [1,2,3] => [2,3,3] => [3,4,3] => [4,4,4] Example 2: Input: nums = [1,1,1] Output: 0 Constraints: n == nums.length 1 <= nums.length <= 10 5 -10 9 <= nums[i] <= 10 9 The answer is guarant

Detailed Explanation

The problem asks us to find the minimum number of moves needed to make all elements in an integer array equal. Each move involves incrementing `n - 1` elements of the array by 1, where `n` is the array's length. The goal is to determine the smallest number of such moves required to achieve a state where all array elements have the same value. The input is an array of integers, and the output is a single integer representing the minimum number of moves.

Solution Approach

The provided solution leverages the insight that incrementing `n - 1` elements is equivalent to decrementing one element. Therefore, instead of thinking about incrementing multiple elements, the problem can be reframed as finding the minimum number of decrements needed to make all elements equal to the minimum value in the array. The solution calculates the sum of differences between each element and the minimum element. This sum represents the total number of moves required.

Step-by-Step Algorithm

Step 1: Find the minimum element in the input array `nums`.
Step 2: Calculate the sum of all elements in the array.
Step 3: Calculate the number of moves by subtracting `len(nums) * minVal` from the sum of the array elements. This formula directly computes the total decrements required to reduce each element to the minimum value.
Step 4: Return the calculated number of moves.

Key Insights

Insight 1: Incrementing `n - 1` elements is equivalent to decrementing only one element. This is the core insight that simplifies the problem.
Insight 2: The target value for all elements to converge to is the minimum value in the original array. Decrementing all other elements until they equal the minimum value is the optimal strategy.
Insight 3: The total number of moves is the sum of the differences between each element and the minimum element.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math.

Companies

Asked at: Coursera, Indeed.

Minimum Moves to Equal Array Elements | LeetCode 453

Medium

Array Math

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

Solution Code

Available in:

class Solution {
    public int minMoves(int[] nums) {
        int minVal = nums[0];
        long sum = 0;
        for (int num : nums) {
            sum += num;
            minVal = Math.min(minVal, num);
        }
        return (int)(sum - (long)nums.length * minVal);
    }
}

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Problem Info

Description

Problem 453: Minimum Moves to Equal Array Elements - Detailed explanation and solution approaches.

Topics

Array Math

Companies

CourseraIndeed

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:453

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

4 steps

1
Step 1: Find the minimum element in the input array `nums`.
2
Step 2: Calculate the sum of all elements in the array.
3
Step 3: Calculate the number of moves by subtracting `len(nums) * minVal` from the sum of the array elements. This formula directly computes the total decrements required to reduce each element to the minimum value.
4
Step 4: Return the calculated number of 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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operations are finding the minimum element and summing the elements of the array. Both of these operations require iterating through the array once, resulting in a linear time complexity O(n). The space used is constant because only a few variables are used to store intermediate values.

📖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

Sorting-based approach

Sort the array in ascending order. Then, iterate through the array and sum the differences between each element and the first element (which is the minimum after sorting).

Time: O(n log n) due to the sorting step.Space: O(1) or O(n) depending on the sorting algorithm used (in-place vs. creating a new sorted array).

Trade-offs: The sorting approach is less efficient than the provided solution due to the O(n log n) time complexity. Use the provided solution's approach unless in-place sorting is absolutely necessary.

✨

Key Insights

1Insight 1: Incrementing `n - 1` elements is equivalent to decrementing only one element. This is the core insight that simplifies the problem.
2Insight 2: The target value for all elements to converge to is the minimum value in the original array. Decrementing all other elements until they equal the minimum value is the optimal strategy.
3Insight 3: The total number of moves is the sum of the differences between each element and the minimum element.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: All elements in the array are the same. In this case, the minimum element will be equal to all other elements, and the sum of differences will be zero, resulting in zero moves, which is correct.
Edge case 2: Large integer values in the input array. The sum of the elements can potentially exceed the maximum value of an integer. Therefore, `long` is used in Java and C++, and `0L` is used in C++ to ensure the sum doesn't overflow. The C version also uses `long long` for the sum.
Edge case 3: Array with only one element. The algorithm will still work correctly. The minimum will be the only element and the sum will also be the same value. The moves required will be the sum - 1 * min_value which equals zero as expected.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to recognize the equivalence between incrementing `n - 1` elements and decrementing one element. This can lead to a much more complex and less efficient solution.

✗Common mistake 2: Integer overflow when calculating the sum of array elements. Using `long` or `long long` to store the sum is crucial, especially when dealing with large arrays and values.

✗Common mistake 3: Assuming that the target value must be the average of the array elements. The target value is always the minimum element.

✅

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 leveling in project management. Imagine each element in the array represents the workload of a team member. Incrementing `n-1` team members effectively means transferring some workload to a single team member. The problem translates to finding the minimum number of workload transfers to equalize the workload across all team members.

🏦Real-world scenario 2: Data normalization in machine learning. Although not a direct application, the concept of adjusting values relative to the minimum or maximum is fundamental in data preprocessing techniques such as min-max scaling.

💡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 2, 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

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

@CodeNinjaJul 13

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

@TechLeadJul 6

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

@DevExpert92Jun 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.

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 "Minimum Moves to Equal Array Elements"?

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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