How do you solve Minimum Deletions to Make Array Beautiful on LeetCode?

Minimum Deletions to Make Array Beautiful (LeetCode #2216) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Adjacent equal elements must be handled to ensure the beautiful condition.

What is the time complexity of Minimum Deletions to Make Array Beautiful?

The optimal solution for Minimum Deletions to Make Array Beautiful runs in O(n) time and uses O(1) space. This complexity is linear because we only traverse the array once, making constant-time checks for adjacent elements.

What is the brute force solution for Minimum Deletions to Make Array Beautiful?

The brute force approach for Minimum Deletions to Make Array Beautiful is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible deletions to see if the resulting array meets the beautiful criteria. This involves generating all possible combinations of deletions, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Deletions to Make Array Beautiful?

Minimum Deletions to Make Array Beautiful uses the following concepts: Array, Stack, Greedy. The recommended patterns to study are: Greedy, Array.

What are the interview tips for Minimum Deletions to Make Array Beautiful?

Always clarify the problem requirements before jumping to solutions. Discuss your thought process and approach with the interviewer. Consider edge cases, such as arrays of odd lengths or all equal elements.

What are common mistakes when solving Minimum Deletions to Make Array Beautiful?

Not checking the final length of the array after processing. Overlooking the condition of adjacent elements at even indices.

#2216

Minimum Deletions to Make Array Beautiful

Medium

ArrayStackGreedyGreedyArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a greedy approach to traverse the array and count the necessary deletions in a single pass. This is efficient because it directly checks adjacent elements without generating subsets.

⚙️

Algorithm

4 steps

1Step 1: Initialize a counter for deletions and a variable to track the last unique element.
2Step 2: Traverse the array, checking pairs of elements at even indices.
3Step 3: If the current element is equal to the last unique element, increment the deletion counter.
4Step 4: If the length of the array is odd after processing, increment the deletion counter to ensure even length.

solution.py9 lines

1def min_deletions_optimal(nums):
2    deletions = 0
3    n = len(nums)
4    for i in range(0, n - 1, 2):
5        if nums[i] == nums[i + 1]:
6            deletions += 1
7    if (n - deletions) % 2 != 0:
8        deletions += 1
9    return deletions

ℹ

Complexity note: This complexity is linear because we only traverse the array once, making constant-time checks for adjacent elements.

1Adjacent equal elements must be handled to ensure the beautiful condition.
2The final length of the array must be even, which may require an additional deletion.

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