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Maximum Median Sum of Subsequences of Size 3
Maximize the sum of medians from subsequences of size 3 by choosing elements wisely from the array.
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Medium · Maximum Median Sum of Subsequences of Size 3 core interview pattern
Answer-first summary
Maximize the sum of medians from subsequences of size 3 by choosing elements wisely from the array.
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To solve this problem, you must pick subsequences of size 3, compute their median, and maximize the sum of all medians. Sorting the array is a helpful first step, as it allows for selecting the largest medians efficiently. The goal is to get the highest possible sum from the medians of the subsequences.
Problem Statement
You are given an integer array nums with a length divisible by 3. In each step, select any three elements from the array, compute their median, and remove them. The median is the middle element of the sorted subsequence of three numbers.
The objective is to maximize the sum of all computed medians after repeating the selection process until the array is empty. Focus on selecting the subsequences strategically to ensure the sum of the medians is as high as possible.
Examples
Example 1
Input: nums = [2,1,3,2,1,3]
Output: 5
Hence, the sum of the medians is 3 + 2 = 5 .
Example 2
Input: nums = [1,1,10,10,10,10]
Output: 20
Hence, the sum of the medians is 10 + 10 = 20 .
Constraints
- 1 <= nums.length <= 5 * 105
- nums.length % 3 == 0
- 1 <= nums[i] <= 109
Solution Approach
Sort the array
Sorting the array in ascending order helps by ensuring you can always select subsequences with higher medians. The median of a sorted triplet is always the middle element, so by focusing on the larger elements, you ensure a higher sum of medians.
Select triplets from the center
After sorting, pick the triplets such that the middle element of each subsequence is maximized. This can be achieved by selecting the elements near the center of the sorted array, as they yield the highest possible median.
Greedy subsequence selection
Using a greedy approach, select the triplets that give the largest median first. By removing the elements with the smallest possible medians last, you maximize the overall sum of medians.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity is determined by the sorting step, which is O(n log n), where n is the length of the input array. After sorting, the subsequence selection can be done in linear time, O(n). The space complexity is O(n) due to the storage required for the sorted array.
What Interviewers Usually Probe
- Understanding of sorting and its impact on selecting medians
- Ability to apply greedy strategies for optimization
- Familiarity with common patterns in subsequence problems
Common Pitfalls or Variants
Common pitfalls
- Incorrect median calculation for subsequences
- Not sorting the array before selection
- Failing to use a greedy approach to maximize the sum of medians
Follow-up variants
- Using a different subsequence size
- Minimizing the sum of medians instead of maximizing
- Allowing duplicates in the array for varied subsequence selections
FAQ
How do I approach the Maximum Median Sum of Subsequences of Size 3 problem?
Start by sorting the array and then select triplets such that the middle element (median) is maximized. Use a greedy strategy to select subsequences.
Why is sorting the array important in this problem?
Sorting the array ensures that you can always select subsequences where the middle element is as large as possible, which maximizes the sum of medians.
What are the best strategies for maximizing the sum of medians?
The best strategy involves selecting triplets from the center of the sorted array and using a greedy approach to pick subsequences with larger medians first.
What is the time complexity of this problem?
The time complexity is dominated by the sorting step, which is O(n log n), where n is the length of the input array.
Can I optimize the space complexity for this problem?
The space complexity is O(n), mainly due to storing the sorted array. To optimize, avoid unnecessary copying of the array.
Solution
Solution 1: Greedy + Sorting
To maximize the sum of medians, we need to select larger elements as medians whenever possible. Since each operation can only select three elements, we can sort the array and then start from index $n / 3$, selecting every other element (skipping one) until the end of the array. This ensures that we select the largest possible medians.
class Solution:
def maximumMedianSum(self, nums: List[int]) -> int:
nums.sort()
return sum(nums[len(nums) // 3 :: 2])