What is the time complexity of Minimum Number of Operations to Make Arrays Similar?

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

What data structures are used to solve Minimum Number of Operations to Make Arrays Similar?

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

What are the key insights for solving Minimum Number of Operations to Make Arrays Similar?

Insight 1: The core observation is that each operation preserves parity. An even number can only be transformed into another even number, and an odd number into another odd number. This allows us to treat even and odd elements separately. Insight 2: Sorting the arrays (or sub-arrays of even and odd numbers) is crucial. After sorting, we can directly map the i-th smallest even number in `nums` to the i-th smallest even number in `target`, and similarly for odd numbers. This minimizes the number of operations.

Which companies ask LeetCode 2449. Minimum Number of Operations to Make Arrays Similar in interviews?

This problem is frequently asked at Walmart Labs.

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Minimum Number of Operations to Make Arrays Similar - LeetCode 2449 Solution

Q: How to solve Minimum Number of Operations to Make Arrays Similar?

The solution leverages the parity preservation property and a greedy approach. It first sorts both `nums` and `target` based on parity (even numbers come before odd numbers), and within each parity group, sorts in ascending order. Then, it calculates the sum of absolute differences between corresponding elements of the sorted arrays. Finally, it divides this sum by 4 to get the minimum number of operations. Dividing by 4 stems from the fact that each operation involves an increment of 2 and a decrement of 2, thus changing the sum of absolute difference by 4.

Minimum Number of Operations to Make Arrays Similar - Complete Solution Guide

Minimum Number of Operations to Make Arrays Similar is LeetCode problem 2449, 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 two positive integer arrays nums and target , of the same length. In one operation, you can choose any two distinct indices i and j where 0 <= i, j < nums.length and: set nums[i] = nums[i] + 2 and set nums[j] = nums[j] - 2 . Two arrays are considered to be similar if the frequency of each element is the same. Return the minimum number of operations required to make nums similar to target . The test cases are generated such that nums can always be similar to target . Example 1: Inpu

Detailed Explanation

The problem asks us to find the minimum number of operations required to make two integer arrays, `nums` and `target`, similar. An array is considered similar if the frequency of each element is the same in both arrays. The allowed operation involves picking two distinct indices `i` and `j` and increasing `nums[i]` by 2 while decreasing `nums[j]` by 2. The constraints guarantee that it's always possible to make the arrays similar.

Solution Approach

The solution leverages the parity preservation property and a greedy approach. It first sorts both `nums` and `target` based on parity (even numbers come before odd numbers), and within each parity group, sorts in ascending order. Then, it calculates the sum of absolute differences between corresponding elements of the sorted arrays. Finally, it divides this sum by 4 to get the minimum number of operations. Dividing by 4 stems from the fact that each operation involves an increment of 2 and a decrement of 2, thus changing the sum of absolute difference by 4.

Step-by-Step Algorithm

Step 1: Sort the `nums` array using a custom key function that prioritizes parity and then the element value. This groups even and odd numbers and sorts them within their respective groups.
Step 2: Sort the `target` array using the same custom key function as in Step 1.
Step 3: Initialize a variable `abs_diff_sum` to 0.
Step 4: Iterate through the sorted arrays, calculating the absolute difference between `nums[i]` and `target[i]` for each index `i`. Add this difference to `abs_diff_sum`.
Step 5: Return `abs_diff_sum // 4`. This gives the minimum number of operations.

Key Insights

Insight 1: The core observation is that each operation preserves parity. An even number can only be transformed into another even number, and an odd number into another odd number. This allows us to treat even and odd elements separately.
Insight 2: Sorting the arrays (or sub-arrays of even and odd numbers) is crucial. After sorting, we can directly map the i-th smallest even number in `nums` to the i-th smallest even number in `target`, and similarly for odd numbers. This minimizes the number of operations.
Insight 3: The total increase and total decrease must balance out. The problem's constraints guarantee this. The total number of operations is related to the sum of absolute differences between corresponding elements after sorting by parity and value.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy, Sorting.

Companies

Asked at: Walmart Labs.

Minimum Number of Operations to Make Arrays Similar | LeetCode 2449

Hard

Array Greedy Sorting

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;

class Solution {
    /**
     * Calculates the minimum number of operations to make nums similar to target.
     *
     * An operation consists of choosing two indices i and j and applying
     * nums[i] += 2 and nums[j] -= 2.
     *
     * The core insights are:
     * 1. The operation preserves the parity of numbers. Thus, odd numbers in nums
     *    can only be transformed into odd numbers in target, and similarly for
     *    even numbers. This means we can solve the problem for odd and even
     *    sub-arrays independently.
     * 2. The sum of elements in the array remains constant. This implies that
     *    the total increase required must equal the total decrease required.
     * 3. To minimize operations, we should transform the i-th smallest odd number
     *    in nums to the i-th smallest odd number in target (and similarly for evens).
     *    This is a greedy strategy that can be proven to be optimal.
     *
     * The algorithm proceeds as follows:
     * 1. Separate both `nums` and `target` into odd and even numbers.
     * 2. Sort these four sub-arrays.
     * 3. Calculate the sum of differences between corresponding elements.
     *    Let `P` be the sum of all positive differences (where `target[i] > nums[i]`).
     *    This `P` represents the total amount of value that needs to be added to
     *    elements of `nums`.
     * 4. Each operation provides an increase of `+2`. Therefore, the total number
     *    of operations is `P / 2`.
     *
     * An equivalent calculation is to find the sum of absolute differences, `S`.
     * It can be shown that `S = 2 * P`. Thus, the number of operations is `S / 4`.
     * This leads to a very concise implementation.
     *
     * A clever way to implement step 1 and 2 together is to sort the original
     * arrays using a custom key `(x % 2, x)`. This groups even numbers before
     * odd numbers, and sorts numbers within each parity group.
     */
    public long makeSimilar(int[] nums, int[] target) {
        // Sort nums and target using a custom comparator
        Arrays.sort(nums);
        Arrays.sort(target);

        Arrays.sort(nums);
        Arrays.sort(target);
        
        List<Integer> numsList = new ArrayList<>();
        for (int num : nums) {
            numsList.add(num);
        }

        List<Integer> targetList = new ArrayList<>();
        for (int num : target) {
            targetList.add(num);
        }

        Collections.sort(numsList, Comparator.comparingInt(x -> x % 2).thenComparingInt(x -> x));
        Collections.sort(targetList, Comparator.comparingInt(x -> x % 2).thenComparingInt(x -> x));


        // Calculate the sum of absolute differences between the paired elements.
        // After sorting by (parity, value), nums[i] should be transformed to target[i].
        // The total sum of positive differences (increments) is half of the sum of
        // all absolute differences.
        // Each operation contributes 2 to the sum of positive differences.
        // Thus, operations = (sum of positive diffs) / 2 = (sum of abs diffs / 2) / 2
        //                  = sum of abs diffs / 4

        long abs_diff_sum = 0;
        for (int i = 0; i < numsList.size(); i++) {
            abs_diff_sum += Math.abs(numsList.get(i) - targetList.get(i));
        }

        return abs_diff_sum / 4;
    }
}

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Problem Info

Description

Problem 2449: Minimum Number of Operations to Make Arrays Similar - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting

Companies

Walmart Labs

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2449

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

5 steps

1
Step 1: Sort the `nums` array using a custom key function that prioritizes parity and then the element value. This groups even and odd numbers and sorts them within their respective groups.
2
Step 2: Sort the `target` array using the same custom key function as in Step 1.
3
Step 3: Initialize a variable `abs_diff_sum` to 0.
4
Step 4: Iterate through the sorted arrays, calculating the absolute difference between `nums[i]` and `target[i]` for each index `i`. Add this difference to `abs_diff_sum`.
5
Step 5: Return `abs_diff_sum // 4`. This gives the minimum number of operations.

💡

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 dominant operation is sorting, which typically has a time complexity of O(n log n). The space complexity depends on the specific sorting algorithm used. If an in-place sorting algorithm is employed, the space complexity is O(1). Otherwise, it will be O(n).

📖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 separate lists for even and odd numbers

Instead of sorting with a custom key, you could create four separate lists: even_nums, odd_nums, even_target, and odd_target. Then, sort each of these lists individually. This is functionally equivalent to the provided solution but might be slightly less concise.

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

Trade-offs: This approach requires more memory due to the creation of additional lists but might be more readable for some developers.

✨

Key Insights

1Insight 1: The core observation is that each operation preserves parity. An even number can only be transformed into another even number, and an odd number into another odd number. This allows us to treat even and odd elements separately.
2Insight 2: Sorting the arrays (or sub-arrays of even and odd numbers) is crucial. After sorting, we can directly map the i-th smallest even number in `nums` to the i-th smallest even number in `target`, and similarly for odd numbers. This minimizes the number of operations.
3Insight 3: The total increase and total decrease must balance out. The problem's constraints guarantee this. The total number of operations is related to the sum of absolute differences between corresponding elements after sorting by parity and value.

📚

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.

Sorting

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

Edge Cases

Edge case 1: `nums` and `target` are already identical. The algorithm correctly handles this case, as the sum of absolute differences will be 0, resulting in 0 operations.
Edge case 2: Arrays contain only even or only odd numbers. The sorting and subsequent calculation work correctly regardless of the specific values in the arrays, as long as the total increase needed equals the total decrease needed (guaranteed by the problem constraints).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the arrays based on parity. If you only sort by value, you might not be matching the correct even numbers with other even numbers, or odd with odd, leading to an incorrect operation count.

✗Common mistake 2: Dividing the sum of absolute differences by 2 instead of 4. Each operation affects two elements, and the changes are +2 and -2. Only half of the absolute differences have to be compensated for and each such operation increases a number by 2. Therefore the result needs to be divided by 4.

✗Common mistake 3: Using integer division when large numbers are involved. Using integer division with very large absolute differences might lead to incorrect results. For Java/C++ it's important to use `long` to prevent overflow.

✅

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 you have two sets of resources (e.g., memory blocks) and you want to redistribute them to match a desired configuration. The problem is analogous to minimizing the number of transfers (operations) needed to achieve the desired allocation, where the amount transferred must be a multiple of a certain unit.

🏦Real-world scenario 2: Balancing network traffic. Suppose you have two networks with varying traffic loads. You want to balance the traffic by transferring data between them. The problem becomes minimizing the number of data transfers, where each transfer involves moving a fixed amount of data between two specific nodes in the networks.

💡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.

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 18, 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 14

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

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

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 "Minimum Number of Operations to Make Arrays Similar"?

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?