What is the time complexity of Minimum Array Changes to Make Differences Equal?

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

What data structures are used to solve Minimum Array Changes to Make Differences Equal?

This problem uses Array, Hash Table, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Array Changes to Make Differences Equal?

Insight 1: The core idea is to iterate through all possible values of X (0 to k) and calculate the minimum number of changes needed to make all pairs (nums[i], nums[n-1-i]) have an absolute difference of X. Insight 2: For each pair (a, b), there are three possibilities: abs(a-b) == X (no change needed), abs(a-b) != X but it can be made equal to X with one change (X <= max(a, b, k - a, k - b)), or abs(a-b) != X and it needs two changes.

Which companies ask LeetCode 3224. Minimum Array Changes to Make Differences Equal in interviews?

This problem is frequently asked at Airbus SE.

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Minimum Array Changes to Make Differences Equal - LeetCode 3224 Solution

Q: How to solve Minimum Array Changes to Make Differences Equal?

The solution calculates the minimum number of changes by iterating through all possible target differences 'X' from 0 to 'k'. For each 'X', it calculates the number of changes required to make the absolute difference of each pair of elements (nums[i], nums[n-1-i]) equal to 'X'. To optimize this calculation, the solution pre-computes the frequency of each absolute difference encountered in the original array and also stores the maximum value achievable (considering changes) for each pair (nums[i], nums[n-1-i]). Using this pre-computed information, the cost (number of changes) for each X is efficiently calculated.

Q: Which companies ask LeetCode 3224. Minimum Array Changes to Make Differences Equal in interviews?

This problem is frequently asked at Airbus SE.

Minimum Array Changes to Make Differences Equal - Complete Solution Guide

Minimum Array Changes to Make Differences Equal is LeetCode problem 3224, 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 of size n where n is even , and an integer k . You can perform some changes on the array, where in one change you can replace any element in the array with any integer in the range from 0 to k . You need to perform some changes (possibly none) such that the final array satisfies the following condition: There exists an integer X such that abs(a[i] - a[n - i - 1]) = X for all (0 <= i < n) . Return the minimum number of changes required to satisfy the above cond

Detailed Explanation

The problem asks us to minimize the number of changes required to an array 'nums' of even length 'n' such that for every index 'i' (0 <= i < n/2), the absolute difference between nums[i] and nums[n - 1 - i] is the same value 'X'. We are allowed to change any element to any value between 0 and 'k' inclusive. The goal is to find the minimum number of changes to achieve this condition.

Solution Approach

The solution calculates the minimum number of changes by iterating through all possible target differences 'X' from 0 to 'k'. For each 'X', it calculates the number of changes required to make the absolute difference of each pair of elements (nums[i], nums[n-1-i]) equal to 'X'. To optimize this calculation, the solution pre-computes the frequency of each absolute difference encountered in the original array and also stores the maximum value achievable (considering changes) for each pair (nums[i], nums[n-1-i]). Using this pre-computed information, the cost (number of changes) for each X is efficiently calculated.

Step-by-Step Algorithm

Step 1: Initialize 'freq' and 'max_diff_counts' arrays of size k+1 to store frequencies and achievable maximum differences respectively. They are initialized to 0.
Step 2: Iterate through the first half of the array (0 to n/2 - 1). For each index 'i', extract the pair (nums[i], nums[n - 1 - i]).
Step 3: Calculate the absolute difference 'd' between the pair (a, b) and increment 'freq[d]'. This counts how many pairs already have a difference of 'd'.
Step 4: Calculate the maximum value achievable within the range [0, k] after changing a or b. This is done by taking the maximum of a, b, k - a, and k - b, and increment 'max_diff_counts[max_diff]'.
Step 5: Calculate the suffix sum of 'max_diff_counts' into 'count_ge_X'. 'count_ge_X[i]' represents the number of pairs whose maximum achievable difference is greater than or equal to 'i'. This is done from k down to 0.
Step 6: Iterate through all possible values of 'X' from 0 to 'k'. For each 'X', calculate the 'cost' which is the total number of changes needed to make all pairs have a difference of 'X'. The initial cost is 'n'. Reduce the cost by 1 for each pair where max_diff >= X (one change needed instead of 2) and by 1 for each pair where d == X (no change needed). Thus, cost = n - count_ge_X[X] - freq[X]
Step 7: Maintain a 'min_cost' variable to store the minimum cost encountered so far. Update it in each iteration if the current 'cost' is smaller.
Step 8: Return 'min_cost'.

Key Insights

Insight 1: The core idea is to iterate through all possible values of X (0 to k) and calculate the minimum number of changes needed to make all pairs (nums[i], nums[n-1-i]) have an absolute difference of X.
Insight 2: For each pair (a, b), there are three possibilities: abs(a-b) == X (no change needed), abs(a-b) != X but it can be made equal to X with one change (X <= max(a, b, k - a, k - b)), or abs(a-b) != X and it needs two changes.
Insight 3: Instead of individually checking for each pair and possible X, pre-compute frequencies and prefix/suffix sums to efficiently count the number of pairs that satisfy the condition of needing 0, 1 or 2 changes for a given X.

Complexity Analysis

Time Complexity: O(k)

Space Complexity: O(k)

Topics

This problem involves: Array, Hash Table, Prefix Sum.

Companies

Asked at: Airbus SE.

Minimum Array Changes to Make Differences Equal | LeetCode 3224

Medium

Array Hash Table Prefix Sum

Time: O(k)Space: O(k)

Solution Code

Available in:

class Solution {
    public int minChanges(int[] nums, int k) {
        int n = nums.length;
        int[] freq = new int[k + 1];
        int[] max_diff_counts = new int[k + 1];

        for (int i = 0; i < n / 2; i++) {
            int a = nums[i];
            int b = nums[n - 1 - i];

            int d = Math.abs(a - b);
            freq[d]++;

            int max_diff = Math.max(Math.max(a, b), Math.max(k - a, k - b));
            max_diff_counts[max_diff]++;
        }

        // The logic is based on calculating the cost for each possible difference X.
        // The cost for a pair (a,b) to have a difference of X is:
        // 0 changes if abs(a-b) == X.
        // 1 change if abs(a-b) != X and X is achievable with one change.
        //   This is possible if X <= max(a, b, k-a, k-b).
        // 2 changes otherwise.
        //
        // Let's derive the total cost C(X) for a fixed X.
        // The baseline cost for each of the n/2 pairs is 2 changes, totaling n changes.
        // We get savings for each pair based on the target X.
        // Save 1 change if X <= max_diff_i.
        // Save another 1 change if X == d_i.
        // So, C(X) = n - (num pairs with max_diff_i >= X) - (num pairs with d_i == X).
        // C(X) = n - count_ge_X[X] - freq[X].

        // To efficiently calculate count_ge_X for all X, we can use a suffix sum.
        int[] count_ge_X = new int[k + 2];
        for (int i = k; i >= 0; i--) {
            count_ge_X[i] = count_ge_X[i + 1] + max_diff_counts[i];
        }

        int min_cost = n;
        for (int X = 0; X <= k; X++) {
            int cost = n - count_ge_X[X] - freq[X];
            min_cost = Math.min(min_cost, cost);
        }

        return min_cost;
    }
}

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Problem Info

Description

Problem 3224: Minimum Array Changes to Make Differences Equal - Detailed explanation and solution approaches.

Topics

Array Hash Table Prefix Sum

Companies

Airbus SE

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3224

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

8 steps

1
Step 1: Initialize 'freq' and 'max_diff_counts' arrays of size k+1 to store frequencies and achievable maximum differences respectively. They are initialized to 0.
2
Step 2: Iterate through the first half of the array (0 to n/2 - 1). For each index 'i', extract the pair (nums[i], nums[n - 1 - i]).
3
Step 3: Calculate the absolute difference 'd' between the pair (a, b) and increment 'freq[d]'. This counts how many pairs already have a difference of 'd'.
4
Step 4: Calculate the maximum value achievable within the range [0, k] after changing a or b. This is done by taking the maximum of a, b, k - a, and k - b, and increment 'max_diff_counts[max_diff]'.
5
Step 5: Calculate the suffix sum of 'max_diff_counts' into 'count_ge_X'. 'count_ge_X[i]' represents the number of pairs whose maximum achievable difference is greater than or equal to 'i'. This is done from k down to 0.
6
Step 6: Iterate through all possible values of 'X' from 0 to 'k'. For each 'X', calculate the 'cost' which is the total number of changes needed to make all pairs have a difference of 'X'. The initial cost is 'n'. Reduce the cost by 1 for each pair where max_diff >= X (one change needed instead of 2) and by 1 for each pair where d == X (no change needed). Thus, cost = n - count_ge_X[X] - freq[X]
7
Step 7: Maintain a 'min_cost' variable to store the minimum cost encountered so far. Update it in each iteration if the current 'cost' is smaller.
8
Step 8: Return 'min_cost'.

💡

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(k)

💾

Space Complexity

O(k)

📝

Why?

The dominant operations are iterating through the range of possible values for the difference X (0 to k) and calculating the suffix sum of `max_diff_counts`. Since all arrays depend only on k and not on n the overall time and space complexity are both O(k).

📖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

Brute-Force Approach

Iterate through all possible values of X from 0 to k and, for each X, iterate through each pair (a, b) in nums. Calculate the number of changes needed to make abs(a-b) == X for that particular pair. Sum up these changes and find the minimum across all X.

Time: O(n*k), where n is the length of nums and k is the upper bound on the values in nums.Space: O(1)

Trade-offs: This approach is less efficient due to the nested loops. The optimized approach described above is much faster because it pre-computes frequencies and prefix sums.

✨

Key Insights

1Insight 1: The core idea is to iterate through all possible values of X (0 to k) and calculate the minimum number of changes needed to make all pairs (nums[i], nums[n-1-i]) have an absolute difference of X.
2Insight 2: For each pair (a, b), there are three possibilities: abs(a-b) == X (no change needed), abs(a-b) != X but it can be made equal to X with one change (X <= max(a, b, k - a, k - b)), or abs(a-b) != X and it needs two changes.
3Insight 3: Instead of individually checking for each pair and possible X, pre-compute frequencies and prefix/suffix sums to efficiently count the number of pairs that satisfy the condition of needing 0, 1 or 2 changes for a given X.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: When n = 2 and nums = [a, b], k = some value. The algorithm correctly handles this case because it calculates the absolute difference and tries all X up to k.
Edge case 2: When all the elements are same in nums, the algorithm correctly sets freq[0] to n/2 and efficiently calculates the min_cost.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the constraint 0 <= nums[i] <= k when calculating the maximum achievable difference for a pair. It's important to consider both changing nums[i] to 0 and changing nums[n-1-i] to k, or vice versa.

✗Common mistake 2: Not calculating the prefix/suffix sum correctly. The correct calculation is critical for achieving the desired time complexity. Specifically when implementing the suffix/prefix sum. Ensure that it is built starting from the largest index.

✅

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: In data analysis, this algorithm could be adapted to minimize the number of data points that need to be changed to satisfy a certain consistency requirement (e.g., making pairs of data entries have a consistent difference).

🏦Real-world scenario 2: In image processing, this algorithm could be adapted to minimize the number of pixel values that need to be changed to achieve a desired contrast or balance between pairs of pixels.

💡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: November 12, 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 13

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

@DevExpert92Jul 6

❓

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 are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 Array Changes to Make Differences Equal"?

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

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