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

Minimum Pair Removal to Sort Array I (LeetCode #3507) 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: Identifying pairs that violate the non-decreasing condition is crucial.

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

The brute force approach for Minimum Pair Removal to Sort Array I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves repeatedly finding the adjacent pair with the minimum sum and replacing it with their sum until the array is non-decreasing. This is straightforward but can be inefficient due to repeated scans of the array. The optimal Optimal Solution approach improves this to O(n).

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

Always clarify the problem statement and constraints. Consider edge cases and how they might affect your solution. Think about the efficiency of your approach and be ready to explain it.

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

Not checking if the array is already sorted before starting operations. Failing to handle edge cases where the array has only one element.

#3507

Minimum Pair Removal to Sort Array I

Easy

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)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution involves a single pass through the array to identify all the necessary pairs to merge, reducing the number of operations by focusing on the overall structure of the array rather than individual pairs.

⚙️

Algorithm

3 steps

1Step 1: Initialize a counter for operations.
2Step 2: Traverse the array and identify pairs that violate the non-decreasing condition.
3Step 3: Merge these pairs and count the operations needed.

solution.py14 lines

1def min_operations(nums):
2    operations = 0
3    while True:
4        merged = False
5        for i in range(len(nums) - 1):
6            if nums[i] > nums[i + 1]:
7                nums[i] += nums[i + 1]
8                del nums[i + 1]
9                operations += 1
10                merged = True
11                break
12        if not merged:
13            break
14    return operations

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once for each operation, and the number of operations is limited by the number of elements in the array.

1Identifying pairs that violate the non-decreasing condition is crucial.
2Merging pairs effectively reduces the number of operations.

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