What is the time complexity of Make Lexicographically Smallest Array by Swapping Elements?

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

What data structures are used to solve Make Lexicographically Smallest Array by Swapping Elements?

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

What are the key insights for solving Make Lexicographically Smallest Array by Swapping Elements?

Insight 1: Elements within the same connected component (where elements can be swapped amongst each other) can be rearranged freely to achieve the lexicographically smallest order within that component. Insight 2: Sorting the elements within each connected component and then placing them back into their original positions within the component results in the lexicographically smallest arrangement for that component.

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Make Lexicographically Smallest Array by Swapping Elements - LeetCode 2948 Solution

Q: How to solve Make Lexicographically Smallest Array by Swapping Elements?

The provided solution identifies connected components within the array. Two indices `i` and `j` are in the same component if `|nums[i] - nums[j]| <= limit`. Once identified, the elements within each component are sorted. Then, the sorted elements are placed back into their original positions within their respective components, resulting in the lexicographically smallest array.

Q: Which companies ask LeetCode 2948. Make Lexicographically Smallest Array by Swapping Elements in interviews?

This problem is frequently asked at Atlassian, PhonePe.

Make Lexicographically Smallest Array by Swapping Elements - Complete Solution Guide

Make Lexicographically Smallest Array by Swapping Elements is LeetCode problem 2948, 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 a 0-indexed array of positive integers nums and a positive integer limit . In one operation, you can choose any two indices i and j and swap nums[i] and nums[j] if |nums[i] - nums[j]| <= limit . Return the lexicographically smallest array that can be obtained by performing the operation any number of times . An array a is lexicographically smaller than an array b if in the first position where a and b differ, array a has an element that is less than the corresponding element in b .

Detailed Explanation

The problem asks us to find the lexicographically smallest array that can be obtained from a given array `nums` by swapping elements. The constraint is that we can only swap two elements `nums[i]` and `nums[j]` if their absolute difference `|nums[i] - nums[j]|` is less than or equal to a given `limit`. The goal is to perform these swaps as many times as needed to achieve the smallest possible array in lexicographical order.

Solution Approach

The provided solution identifies connected components within the array. Two indices `i` and `j` are in the same component if `|nums[i] - nums[j]| <= limit`. Once identified, the elements within each component are sorted. Then, the sorted elements are placed back into their original positions within their respective components, resulting in the lexicographically smallest array.

Step-by-Step Algorithm

Step 1: Create pairs of (value, index) from the input array `nums` and store them in a list/vector. This preserves the original indices of the elements.
Step 2: Sort the pairs based on their values (the first element of the pair). This allows us to easily identify elements that are close enough to be swapped.
Step 3: Iterate through the sorted pairs to identify connected components. A new component starts at the current index `i`. While subsequent elements `j` are within the `limit` difference from the previous element, they belong to the same component.
Step 4: For each identified component, extract the original indices of the elements belonging to the component.
Step 5: Sort these indices to maintain the original order within each component. This will enable placement of the sorted values to the correct location in the result array
Step 6: Assign the sorted values from the sorted pairs to their corresponding positions (original indices) in the result array.
Step 7: Advance the main loop counter `i` to the end of the current component and repeat steps 3-6 until all elements are processed.
Step 8: Return the result array, which now contains the lexicographically smallest arrangement.

Key Insights

Insight 1: Elements within the same connected component (where elements can be swapped amongst each other) can be rearranged freely to achieve the lexicographically smallest order within that component.
Insight 2: Sorting the elements within each connected component and then placing them back into their original positions within the component results in the lexicographically smallest arrangement for that component.
Insight 3: Identifying the connected components is key. Two elements belong to the same component if there exists a path of swappable elements between them (i.e., their absolute difference, or the absolute difference of elements they are connected to, is within the limit).

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Union Find, Sorting.

Companies

Asked at: Atlassian, PhonePe.

Make Lexicographically Smallest Array by Swapping Elements | LeetCode 2948

Medium

Array Union Find Sorting

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

Solution Code

Available in:

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

class Solution {
    public int[] lexicographicallySmallestArray(int[] nums, int limit) {
        int n = nums.length;
        Integer[] indices = new Integer[n];
        for (int i = 0; i < n; i++) {
            indices[i] = i;
        }

        Arrays.sort(indices, Comparator.comparingInt(i -> nums[i]));

        int[] result = new int[n];
        int i = 0;
        while (i < n) {
            int j = i + 1;
            while (j < n && nums[indices[j]].longValue() - nums[indices[j - 1]].longValue() <= limit) {
                j++;
            }

            List<Integer> componentIndices = new ArrayList<>();
            for (int k = i; k < j; k++) {
                componentIndices.add(indices[k]);
            }
            componentIndices.sort(Comparator.naturalOrder());

            for (int k = 0; k < componentIndices.size(); k++) {
                int valToPlace = nums[indices[i + k]];
                int originalIdx = componentIndices.get(k);
                result[originalIdx] = valToPlace;
            }

            i = j;
        }

        return result;
    }
}

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Problem Info

Description

Problem 2948: Make Lexicographically Smallest Array by Swapping Elements - Detailed explanation and solution approaches.

Topics

Array Union Find Sorting

Companies

AtlassianPhonePe

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2948

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: Create pairs of (value, index) from the input array `nums` and store them in a list/vector. This preserves the original indices of the elements.
2
Step 2: Sort the pairs based on their values (the first element of the pair). This allows us to easily identify elements that are close enough to be swapped.
3
Step 3: Iterate through the sorted pairs to identify connected components. A new component starts at the current index `i`. While subsequent elements `j` are within the `limit` difference from the previous element, they belong to the same component.
4
Step 4: For each identified component, extract the original indices of the elements belonging to the component.
5
Step 5: Sort these indices to maintain the original order within each component. This will enable placement of the sorted values to the correct location in the result array
6
Step 6: Assign the sorted values from the sorted pairs to their corresponding positions (original indices) in the result array.
7
Step 7: Advance the main loop counter `i` to the end of the current component and repeat steps 3-6 until all elements are processed.
8
Step 8: Return the result array, which now contains the lexicographically smallest arrangement.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operations are sorting the pairs (O(n log n)) and sorting the component indices (which happens within the loop, but in total it sorts n elements). The outer while loop iterates up to n times. Therefore, the overall time complexity is O(n log n). The space complexity is O(n) because we store the pairs of (value, index), the component indices, and the result array, all of which are proportional to the size of the input array.

📖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

Union Find

Use the Union Find (Disjoint Set Union) data structure to find the connected components. Initially, each element is in its own set. Iterate through all pairs of elements and union them if their absolute difference is within the limit. Then, for each connected component (set), sort the elements and reconstruct the array.

Time: O(n log n) in most implementations, close to O(n) with path compression and union by rank.Space: O(n)

Trade-offs: Union Find can be more efficient in terms of constant factors, especially when the number of unions is large. The provided solution's nested loops could perform redundant checks if there are many overlapping components. Union Find avoids this by efficiently tracking connected components. However, the conceptual complexity of Union Find is slightly higher.

✨

Key Insights

1Insight 1: Elements within the same connected component (where elements can be swapped amongst each other) can be rearranged freely to achieve the lexicographically smallest order within that component.
2Insight 2: Sorting the elements within each connected component and then placing them back into their original positions within the component results in the lexicographically smallest arrangement for that component.
3Insight 3: Identifying the connected components is key. Two elements belong to the same component if there exists a path of swappable elements between them (i.e., their absolute difference, or the absolute difference of elements they are connected to, is within the limit).

📚

Key Terms Explained

For Students

Array

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

Sorting

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

Edge Cases

Edge case 1: Empty array. The code handles this implicitly since the loops won't execute if nums is empty.
Edge case 2: All elements are the same. The sorting will group them together, and the algorithm correctly handles this case by swapping elements with identical values.
Edge case 3: No swappable elements (limit is too small). The algorithm correctly identifies components of size 1, effectively leaving the array unchanged.
Edge case 4: Large limit where all elements are swappable. The algorithm correctly sorts the entire array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly identifying connected components. Failing to check the absolute difference between elements leads to incorrect grouping and wrong results.

✗Common mistake 2: Not sorting the indices of the component before placing the sorted values back into the array. This leads to incorrect placement and a non-lexicographically smallest array.

✗Common mistake 3: Modifying the original input array directly without preserving the original indices. This makes it impossible to reconstruct the array in the correct order.

✗Common mistake 4: Forgetting to sort values within the component. Without this, the smallest elements aren't placed first, violating the lexicographical requirement.

✅

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 network traffic. Imagine a network where nodes can exchange data if their bandwidth capacity is within a certain range. This algorithm can be used to arrange data transfer priorities to minimize latency, resulting in a more efficient network.

🏦Real-world scenario 2: Resource allocation. Consider allocating resources to different tasks based on their resource requirements. Tasks with similar resource requirements can be grouped together and prioritized to optimize overall resource utilization and minimize fragmentation.

💡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: August 8, 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

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

@CodeNinjaJul 13

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

@TechLeadJul 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 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 "Make Lexicographically Smallest Array by Swapping Elements"?

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

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