How do you solve Maximize Greatness of an Array on LeetCode?

Maximize Greatness of an Array (LeetCode #2592) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting the array allows for efficient comparisons to find the next greater element.

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

The optimal solution for Maximize Greatness of an Array runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(1) since we are modifying the array in place.

What is the brute force solution for Maximize Greatness of an Array?

The brute force approach for Maximize Greatness of an Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible permutations of the array and checking how many indices satisfy the condition perm[i] > nums[i]. This is straightforward but inefficient due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximize Greatness of an Array?

Maximize Greatness of an Array uses the following concepts: Array, Two Pointers, Greedy, Sorting. The recommended patterns to study are: Two Pointers, Sorting, Greedy.

What are the interview tips for Maximize Greatness of an Array?

Always consider sorting as a first step for problems involving comparisons. Practice explaining your thought process clearly, as communication is key in interviews. Be prepared to discuss the time and space complexity of your solution.

What are common mistakes when solving Maximize Greatness of an Array?

Failing to sort the array before attempting to find greater elements. Not properly managing the pointers, leading to out-of-bounds errors.

#2592

Maximize Greatness of an Array

Medium

ArrayTwo PointersGreedySortingTwo PointersSortingGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

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Intuition

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

The optimal approach leverages sorting and a two-pointer technique. By sorting the array, we can efficiently find the next greater element for each index, maximizing the count of indices where perm[i] > nums[i].

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Algorithm

4 steps

1Step 1: Sort the input array nums.
2Step 2: Initialize two pointers: one for the sorted array and one for the original array.
3Step 3: Iterate through the original array, and for each element, find the smallest unused element in the sorted array that is greater than the current element.
4Step 4: If found, increment the count and move both pointers forward.

solution.py9 lines

1def maxGreatness(nums):
2    nums.sort()
3    count = 0
4    j = 0
5    for num in nums:
6        if j < len(nums) and num < nums[j]:
7            count += 1
8            j += 1
9    return count

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(1) since we are modifying the array in place.

1Sorting the array allows for efficient comparisons to find the next greater element.
2Using two pointers helps to track the current position in both the original and sorted arrays.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.