How do you solve Sorting Three Groups on LeetCode?

Sorting Three Groups (LeetCode #2826) 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: Understanding the distribution of numbers (1, 2, 3) helps in optimizing the solution.

What is the brute force solution for Sorting Three Groups?

The brute force approach for Sorting Three Groups is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying all possible combinations of removing elements to achieve a non-decreasing order. This means checking every possible way to partition the array into groups of 1s, 2s, and 3s, and counting how many elements need to be removed. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sorting Three Groups?

Sorting Three Groups uses the following concepts: Array, Binary Search, Dynamic Programming. The recommended patterns to study are: Counting Sort, Prefix Sum.

What are the interview tips for Sorting Three Groups?

Always start with a brute-force approach to understand the problem better. Think about how to reduce the problem size or complexity using counting or prefix sums. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Sorting Three Groups?

Overcomplicating the problem by trying to sort instead of counting removals. Not considering edge cases where the array is already sorted.

#2826

Sorting Three Groups

Medium

ArrayBinary SearchDynamic ProgrammingCounting SortPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal approach involves counting how many 1s, 2s, and 3s are in the array and then calculating the minimum removals needed to create a non-decreasing sequence. This is done using prefix sums to keep track of the counts efficiently.

⚙️

Algorithm

3 steps

1Step 1: Count the total number of 1s, 2s, and 3s in the array.
2Step 2: Use prefix sums to track how many 1s and 2s are before each index.
3Step 3: For each possible partition, calculate the number of removals needed and update the minimum removals.

solution.py15 lines

1def min_operations(nums):
2    count1 = nums.count(1)
3    count2 = nums.count(2)
4    count3 = nums.count(3)
5    min_removals = float('inf')
6    prefix1 = 0
7    prefix2 = 0
8    for num in nums:
9        if num == 1:
10            prefix1 += 1
11        elif num == 2:
12            prefix2 += 1
13        removals = (count2 - prefix2) + (count3 - (count2 + count3 - prefix1))
14        min_removals = min(min_removals, removals)
15    return min_removals

ℹ

Complexity note: This complexity is linear because we only pass through the array a couple of times, counting elements and calculating removals.

1Understanding the distribution of numbers (1, 2, 3) helps in optimizing the solution.
2Using prefix sums allows us to efficiently calculate the number of removals needed for each partition.

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