What is the time complexity of Minimum Difference in Sums After Removal of Elements?

The optimal solution for Minimum Difference in Sums After Removal of Elements runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the heap operations for maintaining the top n elements. The space complexity is O(n) because we are storing elements in heaps.

What is the brute force solution for Minimum Difference in Sums After Removal of Elements?

The brute force approach for Minimum Difference in Sums After Removal of Elements is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying every possible combination of removing n elements from the array and calculating the resulting sums of the two parts. This guarantees that we find the minimum difference, but it's inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Difference in Sums After Removal of Elements?

Minimum Difference in Sums After Removal of Elements uses the following concepts: Array, Dynamic Programming, Heap (Priority Queue). The recommended patterns to study are: Heap, Dynamic Programming.

What are the interview tips for Minimum Difference in Sums After Removal of Elements?

Always start with a brute-force solution to understand the problem better. Think about how to optimize your solution by using data structures like heaps. Explain your thought process clearly during the interview to show your understanding.

What are common mistakes when solving Minimum Difference in Sums After Removal of Elements?

Not considering all combinations of removals in the brute-force approach. Failing to properly manage heap sizes in the optimal approach.

#2163

Minimum Difference in Sums After Removal of Elements

Hard

ArrayDynamic ProgrammingHeap (Priority Queue)HeapDynamic Programming

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Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

The optimal approach uses a min-heap and max-heap to efficiently find the minimum sum of n elements and the maximum sum of n elements from the remaining elements after removing n elements. This significantly reduces the time complexity.

⚙️

Algorithm

4 steps

1Step 1: Use a max-heap to keep track of the smallest n elements from the first 2n elements.
2Step 2: Use a min-heap to keep track of the largest n elements from the last 2n elements.
3Step 3: Calculate the sums of the two heaps and find the difference.
4Step 4: Return the minimum difference found.

solution.py23 lines

1# Full working Python code
2import heapq
3
4def minDifference(nums):
5    n = len(nums) // 3
6    max_heap = []  # To store the smallest n elements
7    min_heap = []  # To store the largest n elements
8    
9    # First half processing for max_heap
10    for i in range(2 * n):
11        heapq.heappush(max_heap, nums[i])
12        if len(max_heap) > n:
13            heapq.heappop(max_heap)
14    sum_first = sum(max_heap)
15    
16    # Second half processing for min_heap
17    for i in range(3 * n - 1, n - 1, -1):
18        heapq.heappush(min_heap, -nums[i])
19        if len(min_heap) > n:
20            heapq.heappop(min_heap)
21    sum_second = -sum(min_heap)
22    
23    return sum_first - sum_second

ℹ

Complexity note: The time complexity is O(n log n) due to the heap operations for maintaining the top n elements. The space complexity is O(n) because we are storing elements in heaps.

1The minimum difference can be achieved by carefully selecting which n elements to remove.
2Using heaps allows us to efficiently track the smallest and largest sums needed for the calculation.

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