What is the time complexity of Minimum Operations to Make Array Equal II?

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

What data structures are used to solve Minimum Operations to Make Array Equal II?

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

What are the key insights for solving Minimum Operations to Make Array Equal II?

Insight 1: The net sum of differences between the two arrays must be zero for them to be equalizable using the given operations. If the sum is not zero, it's impossible to make the arrays equal. Insight 2: Each individual difference (nums2[i] - nums1[i]) must be divisible by k. If not, it's impossible to equalize them, because we can only move values in multiples of k.

Which companies ask LeetCode 2541. Minimum Operations to Make Array Equal II in interviews?

This problem is frequently asked at Walmart Labs.

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Minimum Operations to Make Array Equal II - LeetCode 2541 Solution

Q: How to solve Minimum Operations to Make Array Equal II?

The solution iterates through the arrays, calculates the difference between corresponding elements, and checks if the conditions for equalizing the arrays are met. It keeps track of the net sum of differences and the sum of positive differences. If the net sum is not zero or any difference is not divisible by k, it returns -1. Otherwise, it returns the sum of positive differences divided by k, which represents the minimum number of operations.

Minimum Operations to Make Array Equal II - Complete Solution Guide

Minimum Operations to Make Array Equal II is LeetCode problem 2541, 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 two integer arrays nums1 and nums2 of equal length n and an integer k . You can perform the following operation on nums1 : Choose two indexes i and j and increment nums1[i] by k and decrement nums1[j] by k . In other words, nums1[i] = nums1[i] + k and nums1[j] = nums1[j] - k . nums1 is said to be equal to nums2 if for all indices i such that 0 <= i < n , nums1[i] == nums2[i] . Return the minimum number of operations required to make nums1 equal to nums2 . If it is impossible to mak

Detailed Explanation

The problem asks to find the minimum number of operations to make two arrays, `nums1` and `nums2`, equal, given an integer `k`. An operation consists of choosing two indices `i` and `j` and incrementing `nums1[i]` by `k` and decrementing `nums1[j]` by `k`. If it's impossible to make the arrays equal, return -1. The arrays are considered equal if `nums1[i] == nums2[i]` for all indices `i`. The input consists of two integer arrays `nums1` and `nums2` of equal length `n` and an integer `k`. The output is the minimum number of operations, or -1 if the arrays cannot be made equal.

Solution Approach

The solution iterates through the arrays, calculates the difference between corresponding elements, and checks if the conditions for equalizing the arrays are met. It keeps track of the net sum of differences and the sum of positive differences. If the net sum is not zero or any difference is not divisible by k, it returns -1. Otherwise, it returns the sum of positive differences divided by k, which represents the minimum number of operations.

Step-by-Step Algorithm

Step 1: Handle the case where k is 0. If k is 0, check if nums1 and nums2 are equal. If they are, return 0. Otherwise, return -1.
Step 2: Initialize two variables: `net_diff_sum` to store the sum of all differences (nums2[i] - nums1[i]) and `positive_diffs_sum` to store the sum of all positive differences.
Step 3: Iterate through the arrays `nums1` and `nums2` using a loop.
Step 4: In each iteration, calculate the difference `diff = nums2[i] - nums1[i]`.
Step 5: Check if `diff` is divisible by `k`. If not (i.e., `diff % k != 0`), return -1 because it's impossible to make them equal.
Step 6: Update `net_diff_sum` by adding `diff` to it.
Step 7: If `diff` is positive, update `positive_diffs_sum` by adding `diff` to it.
Step 8: After the loop, check if `net_diff_sum` is equal to 0. If not, return -1 because the arrays cannot be made equal.
Step 9: Finally, return `positive_diffs_sum // k`, which represents the minimum number of operations required to make the arrays equal.

Key Insights

Insight 1: The net sum of differences between the two arrays must be zero for them to be equalizable using the given operations. If the sum is not zero, it's impossible to make the arrays equal.
Insight 2: Each individual difference (nums2[i] - nums1[i]) must be divisible by k. If not, it's impossible to equalize them, because we can only move values in multiples of k.
Insight 3: The number of operations is determined by the sum of the positive differences divided by k (or the absolute value of the sum of the negative differences divided by k, which will be the same number). We only need to count the positive differences to determine the number of increment operations.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Greedy.

Companies

Asked at: Walmart Labs.

Minimum Operations to Make Array Equal II | LeetCode 2541

Medium

Array Math Greedy

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

Solution Code

Available in:

class Solution {
    public long minOperations(int[] nums1, int[] nums2, int k) {
        if (k == 0) {
            for (int i = 0; i < nums1.length; i++) {
                if (nums1[i] != nums2[i]) {
                    return -1;
                }
            }
            return 0;
        }

        long net_diff_sum = 0;
        long positive_diffs_sum = 0;

        for (int i = 0; i < nums1.length; i++) {
            int n1 = nums1[i];
            int n2 = nums2[i];
            int diff = n2 - n1;

            if (diff % k != 0) {
                return -1;
            }

            net_diff_sum += diff;

            if (diff > 0) {
                positive_diffs_sum += diff;
            }
        }

        if (net_diff_sum != 0) {
            return -1;
        }

        return positive_diffs_sum / k;
    }
}

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Problem Info

Description

Problem 2541: Minimum Operations to Make Array Equal II - Detailed explanation and solution approaches.

Topics

Array Math Greedy

Companies

Walmart Labs

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2541

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

9 steps

1
Step 1: Handle the case where k is 0. If k is 0, check if nums1 and nums2 are equal. If they are, return 0. Otherwise, return -1.
2
Step 2: Initialize two variables: `net_diff_sum` to store the sum of all differences (nums2[i] - nums1[i]) and `positive_diffs_sum` to store the sum of all positive differences.
3
Step 3: Iterate through the arrays `nums1` and `nums2` using a loop.
4
Step 4: In each iteration, calculate the difference `diff = nums2[i] - nums1[i]`.
5
Step 5: Check if `diff` is divisible by `k`. If not (i.e., `diff % k != 0`), return -1 because it's impossible to make them equal.
6
Step 6: Update `net_diff_sum` by adding `diff` to it.
7
Step 7: If `diff` is positive, update `positive_diffs_sum` by adding `diff` to it.
8
Step 8: After the loop, check if `net_diff_sum` is equal to 0. If not, return -1 because the arrays cannot be made equal.
9
Step 9: Finally, return `positive_diffs_sum // k`, which represents the minimum number of operations required to make the arrays equal.

💡

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 algorithm iterates through the input arrays once, performing constant-time operations at each step. Therefore, the time complexity is linear with respect to the input size. The space complexity is constant because the algorithm uses a fixed number of variables regardless of the input size.

📖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 two separate loops for positive and negative differences

Instead of a single loop, one could use two loops - one to sum positive differences and another to sum negative differences. The absolute values of the sums must be equal for the arrays to be equalizable.

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

Trade-offs: Slightly less efficient than the provided solution due to iterating the array twice, but potentially easier to understand for some.

✨

Key Insights

1Insight 1: The net sum of differences between the two arrays must be zero for them to be equalizable using the given operations. If the sum is not zero, it's impossible to make the arrays equal.
2Insight 2: Each individual difference (nums2[i] - nums1[i]) must be divisible by k. If not, it's impossible to equalize them, because we can only move values in multiples of k.
3Insight 3: The number of operations is determined by the sum of the positive differences divided by k (or the absolute value of the sum of the negative differences divided by k, which will be the same number). We only need to count the positive differences to determine the number of increment operations.

📚

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: `k = 0`. The arrays can only be equalized if they are already equal. The code correctly handles this by directly comparing `nums1` and `nums2` and returning 0 if they are equal, and -1 otherwise.
Edge case 2: The arrays cannot be made equal because the sum of the differences is not zero. The algorithm correctly checks for this by verifying if `net_diff_sum == 0`.
Edge case 3: The arrays cannot be made equal because some differences are not divisible by `k`. The algorithm correctly checks for this by verifying if `diff % k == 0`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking if the differences are divisible by `k`. If a difference is not divisible by `k`, it's impossible to equalize the arrays using the given operation.

✗Common mistake 2: Forgetting to check if the net sum of differences is zero. If the net sum is not zero, the arrays cannot be made equal because there will always be a surplus or deficit.

✗Common mistake 3: Integer overflow when calculating sums. Use `long` data type instead of `int` for accumulators, especially in Java and C++, since problem constraints state that `0 <= nums1[i], nums2[j] <= 10^9` and array length can be up to `10^5`.

✅

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 two servers with differing amounts of resources. The goal is to balance them by moving resources in chunks of size `k`. This problem helps determine if it's possible to balance the servers and the minimum number of transfers required.

🏦Real-world scenario 2: Inventory management. You have two warehouses with different stock levels of a product. You can only move items in batches of `k`. Determining the minimum transfers to equalize inventory levels across warehouses is a real-world application.

💡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: June 22, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 6

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

@CodeNinjaJun 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 Operations to Make Array Equal 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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