How do you solve Minimum Pair Removal to Sort Array II on LeetCode?

Minimum Pair Removal to Sort Array II (LeetCode #3510) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Adjacent pairs can be merged to reduce the array size.

What is the time complexity of Minimum Pair Removal to Sort Array II?

The optimal solution for Minimum Pair Removal to Sort Array II runs in O(n log n) time and uses O(n) space. Using a priority queue allows us to efficiently manage and retrieve the minimum pairs.

What is the brute force solution for Minimum Pair Removal to Sort Array II?

The brute force approach for Minimum Pair Removal to Sort Array II is Brute Force, with O(n²) time complexity and O(1) space complexity. We repeatedly find the adjacent pair with the minimum sum and replace it until the array is sorted. This is inefficient as it may require multiple passes through the array. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Minimum Pair Removal to Sort Array II?

Explain your thought process clearly. Discuss edge cases, like already sorted arrays. Practice similar problems to build intuition.

What are common mistakes when solving Minimum Pair Removal to Sort Array II?

Not updating the priority queue after merging. Forgetting to check if the array is sorted after each operation.

#3510

Minimum Pair Removal to Sort Array II

Hard

ArrayHash TableLinked ListHeap (Priority Queue)SimulationDoubly-Linked ListOrdered SetHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

We can use a priority queue to efficiently find and merge the minimum adjacent pairs, reducing the number of operations needed to sort the array.

⚙️

Algorithm

3 steps

1Step 1: Create a priority queue to store pairs of adjacent elements along with their sums.
2Step 2: Continuously extract the minimum sum pair, merge them, and update the queue until the array is sorted.
3Step 3: Count the number of merge operations performed.

solution.py17 lines

1import heapq
2
3def min_operations(nums):
4    ops = 0
5    pq = []
6    for i in range(len(nums) - 1):
7        heapq.heappush(pq, (nums[i] + nums[i + 1], i))
8    while pq:
9        _, idx = heapq.heappop(pq)
10        nums[idx] += nums[idx + 1]
11        nums.pop(idx + 1)
12        ops += 1
13        if idx > 0:
14            heapq.heappush(pq, (nums[idx - 1] + nums[idx], idx - 1))
15        if idx < len(nums) - 1:
16            heapq.heappush(pq, (nums[idx] + nums[idx + 1], idx))
17    return ops

ℹ

Complexity note: Using a priority queue allows us to efficiently manage and retrieve the minimum pairs.

1Adjacent pairs can be merged to reduce the array size.
2Using a priority queue optimizes the search for minimum pairs.

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