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

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

What data structures are used to solve Minimum Operations to Make Median of Array Equal to K?

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

What are the key insights for solving Minimum Operations to Make Median of Array Equal to K?

Insight 1: The problem requires changing elements of the array to make the median equal to `k`. This implies sorting the array first to easily identify the median. Insight 2: The most efficient approach is to minimize the changes made to elements closest to the median, focusing adjustments around the current median index.

Which companies ask LeetCode 3107. Minimum Operations to Make Median of Array Equal to K in interviews?

This problem is frequently asked at IBM.

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Minimum Operations to Make Median of Array Equal to K - LeetCode 3107 Solution

Minimum Operations to Make Median of Array Equal to K - Complete Solution Guide

Minimum Operations to Make Median of Array Equal to K is LeetCode problem 3107, 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 an integer array nums and a non-negative integer k . In one operation, you can increase or decrease any element by 1. Return the minimum number of operations needed to make the median of nums equal to k . The median of an array is defined as the middle element of the array when it is sorted in non-decreasing order. If there are two choices for a median, the larger of the two values is taken. Example 1: Input: nums = [2,5,6,8,5], k = 4 Output: 2 Explanation: We can subtract one from

Detailed Explanation

The problem asks us to find the minimum number of operations required to make the median of a given array `nums` equal to a target value `k`. An operation consists of either increasing or decreasing any element of the array by 1. The median is defined as the middle element of the sorted array. If the array has an even length, the larger of the two middle elements is considered the median.

Solution Approach

The provided solution sorts the input array `nums` and finds the median index. Then, it compares the value at the median index with the target value `k`. If the median is greater than `k`, it iterates backward from the median index, increasing the number of operations by the difference between each element greater than `k` and `k`. If the median is less than `k`, it iterates forward from the median index, increasing the number of operations by the difference between `k` and each element less than `k`. This approach is based on the fact that changing elements closer to the median affects the median more directly than changing elements far away.

Step-by-Step Algorithm

Step 1: Sort the input array `nums` in non-decreasing order.
Step 2: Calculate the index of the median element: `median_index = n // 2`, where `n` is the length of the array.
Step 3: Initialize a variable `operations` to 0. This variable will store the minimum number of operations required.
Step 4: Compare `nums[median_index]` with `k`. If `nums[median_index] > k`, iterate from `median_index` down to 0. For each element `nums[i]` greater than `k`, add `nums[i] - k` to `operations`. Stop when an element is found that is not greater than `k`.
Step 5: If `nums[median_index] < k`, iterate from `median_index` up to `n - 1`. For each element `nums[i]` less than `k`, add `k - nums[i]` to `operations`. Stop when an element is found that is not less than `k`.
Step 6: Return the value of `operations`.

Key Insights

Insight 1: The problem requires changing elements of the array to make the median equal to `k`. This implies sorting the array first to easily identify the median.
Insight 2: The most efficient approach is to minimize the changes made to elements closest to the median, focusing adjustments around the current median index.
Insight 3: The number of operations needed is simply the absolute difference between the elements that need adjustment and the target value `k`.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy, Sorting.

Companies

Asked at: IBM.

Minimum Operations to Make Median of Array Equal to K | LeetCode 3107

Medium

Array Greedy Sorting

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public long minOperationsToMakeMedianK(int[] nums, int k) {
        Arrays.sort(nums);
        int n = nums.length;
        int median_index = n / 2;
        long operations = 0;

        if (nums[median_index] > k) {
            for (int i = median_index; i >= 0; i--) {
                if (nums[i] > k) {
                    operations += nums[i] - k;
                } else {
                    break;
                }
            }
        } else if (nums[median_index] < k) {
            for (int i = median_index; i < n; i++) {
                if (nums[i] < k) {
                    operations += k - nums[i];
                } else {
                    break;
                }
            }
        }

        return operations;
    }
}

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Problem Info

Description

Problem 3107: Minimum Operations to Make Median of Array Equal to K - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting

Companies

IBM

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3107

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 non-decreasing order.
2
Step 2: Calculate the index of the median element: `median_index = n // 2`, where `n` is the length of the array.
3
Step 3: Initialize a variable `operations` to 0. This variable will store the minimum number of operations required.
4
Step 4: Compare `nums[median_index]` with `k`. If `nums[median_index] > k`, iterate from `median_index` down to 0. For each element `nums[i]` greater than `k`, add `nums[i] - k` to `operations`. Stop when an element is found that is not greater than `k`.
5
Step 5: If `nums[median_index] < k`, iterate from `median_index` up to `n - 1`. For each element `nums[i]` less than `k`, add `k - nums[i]` to `operations`. Stop when an element is found that is not less than `k`.
6
Step 6: Return the value 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 time complexity is dominated by the sorting step, which takes O(n log n) time using efficient sorting algorithms like quicksort or mergesort. The subsequent iterations take at most O(n) time, but this is overshadowed by the sorting time. The space complexity is O(1) because the sorting is done in-place (or uses a constant amount of extra space) and only a few constant extra variables are used.

📖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 Binary Search to Find Elements to Adjust

Instead of iterating linearly to find elements to adjust, we could potentially use binary search to find the first element greater than `k` (when the median is greater than `k`) or the first element less than `k` (when the median is less than `k`). This could improve performance in some cases, but the overall time complexity would remain O(n log n) due to the sorting step.

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

Trade-offs: The potential improvement from binary search might be marginal, and the implementation might be slightly more complex. This approach might be beneficial if the array is extremely large and sorted frequently, as it would reduce the constant factor associated with the linear search.

✨

Key Insights

1Insight 1: The problem requires changing elements of the array to make the median equal to `k`. This implies sorting the array first to easily identify the median.
2Insight 2: The most efficient approach is to minimize the changes made to elements closest to the median, focusing adjustments around the current median index.
3Insight 3: The number of operations needed is simply the absolute difference between the elements that need adjustment and the target value `k`.

📚

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: The input array `nums` is already sorted and the median is equal to `k`. In this case, the solution correctly returns 0 because no operations are needed.
Edge case 2: The input array `nums` contains duplicate values. The solution still works correctly because sorting places the elements in the right order, and the comparison with `k` ensures the correct number of operations is calculated.
Edge case 3: An empty or null input array. While the problem statement doesn't specify this case directly, good code should handle this gracefully, ideally by returning 0 or throwing an exception with descriptive error messages. The given solutions implicitly handle an array with only one element correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the array first. The median cannot be determined correctly if the array is not sorted, leading to incorrect results.

✗Common mistake 2: Not considering the case where the median is already equal to `k`. This will lead to unnecessary iterations and potentially incorrect results, even if the array is sorted.

✗Common mistake 3: Incorrectly calculating the median index for even-length arrays. The problem statement specifies taking the *larger* of the two middle elements in such cases, and the code should correctly implement that.

✗Common mistake 4: Integer overflow when calculating `nums[i] - k` or `k - nums[i]`. Using `long long` for intermediate calculations (as done in the C and C++ versions) is crucial to handle cases where the difference is very large.

✅

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 network bandwidth or CPU resources to different applications. You might want to adjust the resources allocated to each application to achieve a target median resource allocation, minimizing the number of adjustments needed.

🏦Real-world scenario 2: Financial portfolio management. You might have a portfolio of investments, and you want to adjust the holdings to achieve a target median asset allocation. This algorithm can help minimize the number of trades needed to reach the desired allocation.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: July 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

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

@TechLeadJun 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.

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 Median of Array Equal to K"?

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

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