What is the time complexity of Maximize Greatness of an Array?

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

What data structures are used to solve Maximize Greatness of an Array?

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

What are the key insights for solving Maximize Greatness of an Array?

Insight 1: Sorting the input array `nums` allows us to easily compare elements and determine how many elements can be placed in `perm` such that `perm[i] > nums[i]`. Insight 2: A greedy approach is optimal: for each element in the sorted `nums`, we want to find the smallest element in the remaining elements of `nums` that is greater than it. By doing this, we maximize the chance of finding a greater element and preserve the larger elements for potential matches later.

Which companies ask LeetCode 2592. Maximize Greatness of an Array in interviews?

This problem is frequently asked at BlackRock, Salesforce, Twilio, WeRide.

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Maximize Greatness of an Array - LeetCode 2592 Solution

Maximize Greatness of an Array - Complete Solution Guide

Maximize Greatness of an Array is LeetCode problem 2592, 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 integer array nums . You are allowed to permute nums into a new array perm of your choosing. We define the greatness of nums be the number of indices 0 <= i < nums.length for which perm[i] > nums[i] . Return the maximum possible greatness you can achieve after permuting nums . Example 1: Input: nums = [1,3,5,2,1,3,1] Output: 4 Explanation: One of the optimal rearrangements is perm = [2,5,1,3,3,1,1]. At indices = 0, 1, 3, and 4, perm[i] > nums[i]. Hence, we return 4. Exa

Detailed Explanation

The problem asks us to find the maximum possible 'greatness' of an array `nums` after permuting its elements into a new array `perm`. The greatness is defined as the number of indices `i` where `perm[i]` is greater than `nums[i]`. In essence, we need to rearrange the elements of `nums` to maximize the number of times an element in the permuted array is larger than the corresponding element in the original array. The input is an array of integers `nums`, and the output is a single integer representing the maximum possible greatness.

Solution Approach

The provided solution uses a greedy approach. First, the input array `nums` is sorted in ascending order. This sorting allows for efficient comparison and matching of elements. We then iterate through the sorted `nums`. For each element `num` in `nums`, we check if there is an element to the right that's strictly greater. The `greatness` variable keeps track of the index to consider for possible matches. If `num > nums[greatness]`, we increment `greatness`, indicating that we have found a valid match, meaning `nums[greatness]` from the original array can be positioned at index `i` to increase 'greatness'. This approach maximizes the number of indices where `perm[i] > nums[i]`.

Step-by-Step Algorithm

Step 1: Sort the input array `nums` in ascending order.
Step 2: Initialize a variable `greatness` to 0. This variable will store the count of indices where `perm[i] > nums[i]`.
Step 3: Iterate through the sorted array `nums` from the beginning.
Step 4: In each iteration, compare the current element `nums[i]` with `nums[greatness]` (where `i` is the index of the current element being examined).
Step 5: If `nums[i]` is greater than `nums[greatness]`, increment `greatness`. This implies that the element pointed by `greatness` can be positioned at index i in a different permutation such that the perm[i] > nums[i].
Step 6: After iterating through all the elements of `nums`, return the final value of `greatness`.

Key Insights

Insight 1: Sorting the input array `nums` allows us to easily compare elements and determine how many elements can be placed in `perm` such that `perm[i] > nums[i]`.
Insight 2: A greedy approach is optimal: for each element in the sorted `nums`, we want to find the smallest element in the remaining elements of `nums` that is greater than it. By doing this, we maximize the chance of finding a greater element and preserve the larger elements for potential matches later.
Insight 3: The problem essentially boils down to finding, for each element in the sorted array, if there exists an element to its right (after potential rearrangements) that is greater than it.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers, Greedy, Sorting.

Companies

Asked at: BlackRock, Salesforce, Twilio, WeRide.

Maximize Greatness of an Array | LeetCode 2592

Medium

Array Two Pointers Greedy Sorting

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int maximizeGreatness(int[] nums) {
        Arrays.sort(nums);
        int greatness = 0;
        for (int num : nums) {
            if (num > nums[greatness]) {
                greatness += 1;
            }
        }
        return greatness;
    }
}

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Problem Info

Description

Problem 2592: Maximize Greatness of an Array - Detailed explanation and solution approaches.

Topics

Array Two Pointers Greedy Sorting

Companies

BlackRockSalesforceTwilioWeRide

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:2592

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 ascending order.
2
Step 2: Initialize a variable `greatness` to 0. This variable will store the count of indices where `perm[i] > nums[i]`.
3
Step 3: Iterate through the sorted array `nums` from the beginning.
4
Step 4: In each iteration, compare the current element `nums[i]` with `nums[greatness]` (where `i` is the index of the current element being examined).
5
Step 5: If `nums[i]` is greater than `nums[greatness]`, increment `greatness`. This implies that the element pointed by `greatness` can be positioned at index i in a different permutation such that the perm[i] > nums[i].
6
Step 6: After iterating through all the elements of `nums`, return the final value of `greatness`.

💡

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 standard sorting algorithms like merge sort or quicksort (in Java/Python/C++). The subsequent iteration through the sorted array takes O(n) time, but this is less significant than the sorting. The space complexity is O(1) because the sorting is done in-place, and we only use a few extra variables (like `greatness`) which take constant space.

📖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

For each element in the sorted array, perform a binary search for the smallest element greater than it in the remaining array. This approach can also find the `greatness` but might not be as efficient in practice due to the overhead of repeatedly calling binary search. Note that using binary search after removing elements would require a more complex data structure.

Time: O(n log n) on average, where n is the length of the array.Space: O(n) depending on the implementation of the search.

Trade-offs: Sorting is still needed upfront. Binary search would need to ensure the current item is skipped in the search, requiring more book-keeping. Sorting and single pass, while simpler, is generally more efficient in practice here. This approach becomes advantageous if you only want to compute greatness for a subset of values.

✨

Key Insights

1Insight 1: Sorting the input array `nums` allows us to easily compare elements and determine how many elements can be placed in `perm` such that `perm[i] > nums[i]`.
2Insight 2: A greedy approach is optimal: for each element in the sorted `nums`, we want to find the smallest element in the remaining elements of `nums` that is greater than it. By doing this, we maximize the chance of finding a greater element and preserve the larger elements for potential matches later.
3Insight 3: The problem essentially boils down to finding, for each element in the sorted array, if there exists an element to its right (after potential rearrangements) that is greater than it.

📚

Key Terms Explained

For Students

Array

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

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` is empty. The code will handle this gracefully because sorting an empty array has no impact, and the loop will not execute, returning a `greatness` of 0, which is correct.
Edge case 2: `nums` contains only equal elements (e.g., [5, 5, 5, 5]). In this case, no element can be greater than any other element in `nums`, so the `greatness` will remain 0, which is the correct answer.
Edge case 3: `nums` is already sorted in descending order. The code sorts the array, making it ascending. The `greatness` will be computed based on the sorted array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the array. Without sorting, it's difficult to find the optimal permutation efficiently. Sorting allows us to use a greedy approach.

✗Common mistake 2: Incorrectly implementing the greedy strategy. The key is to find the *smallest* element greater than the current element, ensuring we maximize the overall greatness. The existing code's usage of the `greatness` index ensures the smallest viable number is always tracked to check if the current number can exceed it.

✗Common mistake 3: Not handling edge cases where the array is empty or contains only equal elements. The given solution already correctly handles these cases implicitly.

✅

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 tasks to workers, where each task has a required skill level, and each worker has a skill level. The goal is to maximize the number of tasks assigned to workers whose skill level is *greater* than the task's requirement. This problem is analogous to maximizing greatness.

🏦Real-world scenario 2: Performance evaluation. Consider ranking employees based on their performance scores. The goal might be to determine how many employees have a performance score exceeding the average, median, or a specific threshold. This involves comparing scores and finding the number of elements that are 'greater' than a certain value, which aligns with the concept of greatness.

💡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: August 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

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

❓

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.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

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 "Maximize Greatness of an Array"?

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

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