What is the time complexity of Minimum Moves to Make Array Complementary?

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

What data structures are used to solve Minimum Moves to Make Array Complementary?

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 Moves to Make Array Complementary?

Insight 1: The key idea is to iterate through pairs of elements (nums[i], nums[n-1-i]) and determine the number of moves required to make their sum equal to a common target sum. Instead of trying all possible combinations and brute-forcing, we aim to efficiently track the changes using a difference array. Insight 2: The difference array is crucial. It helps calculate the 'savings' (reduction in moves) gained by making the sum of pairs equal to a particular target sum. The changes are tracked using incremental updates on the difference array and a final prefix-sum operation to derive the savings.

Which companies ask LeetCode 1674. Minimum Moves to Make Array Complementary in interviews?

This problem is frequently asked at CureFit.

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Minimum Moves to Make Array Complementary - LeetCode 1674 Solution

Q: What is the space complexity of Minimum Moves to Make Array Complementary?

The space complexity is O(limit).

Minimum Moves to Make Array Complementary - Complete Solution Guide

Minimum Moves to Make Array Complementary is LeetCode problem 1674, 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 even length n and an integer limit . In one move, you can replace any integer from nums with another integer between 1 and limit , inclusive. The array nums is complementary if for all indices i ( 0-indexed ), nums[i] + nums[n - 1 - i] equals the same number. For example, the array [1,2,3,4] is complementary because for all indices i , nums[i] + nums[n - 1 - i] = 5 . Return the minimum number of moves required to make nums complementary . Example 1: Input:

Detailed Explanation

The problem requires you to find the minimum number of moves to make an array 'nums' complementary. An array is complementary if for every index 'i', nums[i] + nums[n - 1 - i] equals the same value. You are given an integer array 'nums' of even length 'n', and an integer 'limit'. In one move, you can replace any integer from 'nums' with another integer between 1 and 'limit' (inclusive). The goal is to determine the fewest number of such moves required to make 'nums' complementary.

Solution Approach

The solution employs a difference array technique to efficiently calculate the minimum number of moves. First, iterate through each pair of elements in the array (nums[i], nums[n-1-i]). For each pair, analyze the possible moves and their impact on the overall moves required to make the array complementary. A difference array ('diff') is used to keep track of the number of moves saved for each potential target sum. By updating the diff array based on the properties of each pair and then computing its prefix sum, we can find the maximum number of total moves saved.

Step-by-Step Algorithm

Step 1: Initialize a difference array 'diff' of size 2 * limit + 2 with all elements set to 0. This array will store the changes in the number of moves required for each possible target sum (from 2 to 2 * limit).
Step 2: Iterate through the first half of the array (from i = 0 to n // 2 - 1) to process each pair (nums[i], nums[n - 1 - i]).
Step 3: For each pair (a, b), determine the minimum (x) and maximum (y) of the two numbers. Calculate their sum (s = a + b).
Step 4: Update the difference array to reflect the savings in moves: - Increasing diff[x + 1] by 1 means that any target sum greater than or equal to x + 1 will save one move. - Decreasing diff[y + limit + 1] by 1 means that the savings of one move will cease for target sums greater than y + limit. - Increasing diff[s] by 1 signifies that making the target sum equal to 's' will save one additional move. - Decreasing diff[s + 1] by 1 means that the additional saving is only applicable for target sum 's' and doesn't extend beyond it.
Step 5: Calculate the prefix sum of the difference array to determine the actual savings for each possible target sum. Iterate from 2 to 2 * limit, accumulating the values in 'diff' to compute the current savings.
Step 6: Keep track of the maximum savings encountered during the prefix sum calculation. This maximum savings represents the largest possible reduction in the total number of moves.
Step 7: Calculate the minimum number of moves required by subtracting the maximum savings from 'n' (which equals 2 * n/2 moves if each pair requires two moves initially) or alternatively n/2 * 2 - max_savings or n - max_savings.

Key Insights

Insight 1: The key idea is to iterate through pairs of elements (nums[i], nums[n-1-i]) and determine the number of moves required to make their sum equal to a common target sum. Instead of trying all possible combinations and brute-forcing, we aim to efficiently track the changes using a difference array.
Insight 2: The difference array is crucial. It helps calculate the 'savings' (reduction in moves) gained by making the sum of pairs equal to a particular target sum. The changes are tracked using incremental updates on the difference array and a final prefix-sum operation to derive the savings.
Insight 3: Consider the range of possible target sums. The minimum possible sum of a pair is 2 (1 + 1), and the maximum is 2 * limit. Thus, the difference array is initialized to have a size of 2 * limit + 2 to store these values.

Complexity Analysis

Time Complexity: O(n + limit)

Space Complexity: O(limit)

Topics

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

Companies

Asked at: CureFit.

Minimum Moves to Make Array Complementary | LeetCode 1674

Medium

Array Hash Table Prefix Sum

Time: O(n + limit)Space: O(limit)

Solution Code

Available in:

class Solution {
    public int minMoves(int[] nums, int limit) {
        int n = nums.length;
        int[] diff = new int[2 * limit + 2];

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

            int x = Math.min(a, b);
            int y = Math.max(a, b);
            int s = a + b;

            // For a pair (x,y), the savings compared to 2 moves are:
            // T in [x+1, y+limit] gives a saving of 1 move.
            // T = s gives an additional saving of 1 move.

            // Contribution to difference array for saving of 1
            // for range [x+1, y+limit]
            diff[x + 1]++;
            diff[y + limit + 1]--;

            // Contribution for additional saving of 1 at T=s
            diff[s]++;
            diff[s + 1]--;
        }

        int max_savings = 0;
        int current_savings = 0;
        // Iterate through all possible target sums T from 2 to 2*limit
        for (int t = 2; t <= 2 * limit; t++) {
            current_savings += diff[t];
            max_savings = Math.max(max_savings, current_savings);
        }

        // The default number of moves for n/2 pairs is n (2 moves per pair).
        // The minimum number of moves is this default minus the maximum savings we can achieve.
        return n - max_savings;
    }
}

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Problem Info

Description

Problem 1674: Minimum Moves to Make Array Complementary - Detailed explanation and solution approaches.

Topics

Array Hash Table Prefix Sum

Companies

CureFit

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1674

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

7 steps

1
Step 1: Initialize a difference array 'diff' of size 2 * limit + 2 with all elements set to 0. This array will store the changes in the number of moves required for each possible target sum (from 2 to 2 * limit).
2
Step 2: Iterate through the first half of the array (from i = 0 to n // 2 - 1) to process each pair (nums[i], nums[n - 1 - i]).
3
Step 3: For each pair (a, b), determine the minimum (x) and maximum (y) of the two numbers. Calculate their sum (s = a + b).
4
Step 4: Update the difference array to reflect the savings in moves: - Increasing diff[x + 1] by 1 means that any target sum greater than or equal to x + 1 will save one move. - Decreasing diff[y + limit + 1] by 1 means that the savings of one move will cease for target sums greater than y + limit. - Increasing diff[s] by 1 signifies that making the target sum equal to 's' will save one additional move. - Decreasing diff[s + 1] by 1 means that the additional saving is only applicable for target sum 's' and doesn't extend beyond it.
5
Step 5: Calculate the prefix sum of the difference array to determine the actual savings for each possible target sum. Iterate from 2 to 2 * limit, accumulating the values in 'diff' to compute the current savings.
6
Step 6: Keep track of the maximum savings encountered during the prefix sum calculation. This maximum savings represents the largest possible reduction in the total number of moves.
7
Step 7: Calculate the minimum number of moves required by subtracting the maximum savings from 'n' (which equals 2 * n/2 moves if each pair requires two moves initially) or alternatively n/2 * 2 - max_savings or n - max_savings.

💡

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 + limit)

💾

Space Complexity

O(limit)

📝

Why?

The time complexity is derived from the loops that iterate n/2 times and 2*limit times, resulting in O(n/2 + 2*limit). Since constants are dropped and n and limit are treated equally from a complexity perspective, it simplifies to O(n + limit). The space complexity is O(limit) because the difference array's size is directly proportional to the limit.

📖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 target sums (from 2 to 2 * limit) and for each sum, calculate the number of moves needed to make all pairs add up to that sum. Keep track of the minimum moves.

Time: O(n * limit^2)Space: O(1)

Trade-offs: This approach is very inefficient due to nested loops (iterating through possible target sums and then through all pairs). It's useful only for very small input sizes. The difference array approach is significantly more efficient.

✨

Key Insights

1Insight 1: The key idea is to iterate through pairs of elements (nums[i], nums[n-1-i]) and determine the number of moves required to make their sum equal to a common target sum. Instead of trying all possible combinations and brute-forcing, we aim to efficiently track the changes using a difference array.
2Insight 2: The difference array is crucial. It helps calculate the 'savings' (reduction in moves) gained by making the sum of pairs equal to a particular target sum. The changes are tracked using incremental updates on the difference array and a final prefix-sum operation to derive the savings.
3Insight 3: Consider the range of possible target sums. The minimum possible sum of a pair is 2 (1 + 1), and the maximum is 2 * limit. Thus, the difference array is initialized to have a size of 2 * limit + 2 to store these values.

📚

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 the array is already complementary (all pairs sum to the same value). In this case, the 'max_savings' will be n/2 and n - max_savings will correctly return 0.
Edge case 2: When the limit is small and the nums[i] values are also small. The difference array will still be of size 2*limit + 2 and the calculations will still be valid.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the savings and applying them to the difference array. Understanding the logic behind the diff[x+1]++, diff[y+limit+1]--, diff[s]++, diff[s+1]-- updates is crucial.

✗Common mistake 2: Failing to properly initialize the difference array or miscalculating the size of the array. The array should be of size 2 * limit + 2 to accommodate all possible target sums.

✅

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: Optimizing resource allocation. Imagine you need to pair up tasks to be processed by two machines, and you want to minimize the number of adjustments needed so that the processing time for each pair is roughly equal. The problem can be modeled using the 'complementary pairs' logic.

🏦Real-world scenario 2: Data balancing in distributed systems. When distributing data across multiple servers, you might want to ensure a balanced load. This could involve making complementary pairs with approximately equal data size so each machine performs roughly the same amount of work. The problem could be framed using the concepts in this question.

💡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: July 25, 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

❓

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 Moves to Make Array Complementary"?

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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