How do you solve Subsequences with a Unique Middle Mode I on LeetCode?

Subsequences with a Unique Middle Mode I (LeetCode #3395) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The middle element must be the unique mode, meaning it must appear more than any other element in the subsequence.

What is the time complexity of Subsequences with a Unique Middle Mode I?

The optimal solution for Subsequences with a Unique Middle Mode I runs in O(n) time and uses O(n) space. The optimal solution runs in O(n) because we only traverse the array a couple of times to build frequency counts and calculate combinations.

What is the brute force solution for Subsequences with a Unique Middle Mode I?

The brute force approach for Subsequences with a Unique Middle Mode I is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible subsequences of size 5 from the array and check if each has a unique middle mode. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Subsequences with a Unique Middle Mode I?

Subsequences with a Unique Middle Mode I uses the following concepts: Array, Hash Table, Math, Combinatorics. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Subsequences with a Unique Middle Mode I?

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases and how they might affect your solution. Practice explaining your thought process as you code; it helps solidify your understanding.

What are common mistakes when solving Subsequences with a Unique Middle Mode I?

Failing to check if the middle element is the unique mode before counting combinations. Not considering edge cases where there are not enough elements to form a valid subsequence.

#3395

Subsequences with a Unique Middle Mode I

Hard

ArrayHash TableMathCombinatoricsHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Instead of generating all combinations, we can directly count valid subsequences by iterating through the array and using frequency counts to determine valid configurations around each potential middle mode.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each number in the array.
2Step 2: For each unique number, consider it as the middle mode and calculate how many valid combinations can be formed with it as the middle element.
3Step 3: Use combinatorial mathematics to count how many ways we can select elements on the left and right of the middle mode.

solution.py18 lines

1# Full working Python code
2from collections import Counter
3from math import comb
4
5def count_unique_middle_mode_subsequences(nums):
6    mod = 10**9 + 7
7    freq = Counter(nums)
8    count = 0
9    for mid in freq:
10        if freq[mid] < 3:
11            continue
12        left_count = sum(v for k, v in freq.items() if k < mid)
13        right_count = sum(v for k, v in freq.items() if k > mid)
14        mid_count = freq[mid]
15        # Choose 2 from mid_count and 1 from left and right
16        count += comb(mid_count, 3) * comb(left_count, 1) * comb(right_count, 1)
17        count %= mod
18    return count

ℹ

Complexity note: The optimal solution runs in O(n) because we only traverse the array a couple of times to build frequency counts and calculate combinations.

1The middle element must be the unique mode, meaning it must appear more than any other element in the subsequence.
2Using combinatorial counting allows us to efficiently calculate the number of valid subsequences without generating them.

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