#3507

Easy

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Minimum Pair Removal to Sort Array I

This problem asks for the minimum number of operations to make an array non-decreasing by removing pairs of elements.

array hash table linked list heap priority queue simulation

Problem Statement

You are given an array nums. You can repeatedly remove one pair of elements from the array until it becomes non-decreasing. A pair of elements is removed if one of them is smaller than the other, but only the pair’s position is considered, not their values themselves. Your task is to return the minimum number of operations required to make the array non-decreasing.

An array is considered non-decreasing if for every i, nums[i] <= nums[i+1]. If the array is already non-decreasing, no operations are required. You can only remove one pair of elements in each operation. A solution that involves scanning the array and leveraging hash lookups will help minimize the operations.

Examples

Example 1

Input: nums = [5,2,3,1]

Output: 2

The array nums became non-decreasing in two operations.

Example 2

Input: nums = [1,2,2]

Output: 0

The array nums is already sorted.

Constraints

1 <= nums.length <= 50
-1000 <= nums[i] <= 1000

Solution Approach

Array Scanning

Iterate through the array from left to right, checking the order of elements. For each element, find the largest subset of consecutive elements that can be removed without breaking the non-decreasing property.

Hash Lookup Optimization

Use a hash set or table to keep track of which pairs can be removed efficiently. This avoids repetitive checks and reduces the computational cost of each step.

Greedy Pair Removal

Perform greedy pair removals by continuously removing the smallest removable pair in each operation. This ensures the minimum number of operations by focusing on the most impactful removals.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is driven by the scanning of the array and the hash lookup process. In the worst case, it will be O(n) where n is the length of the array. Space complexity depends on the data structure used for tracking pairs, but it typically remains O(n) as well.

What Interviewers Usually Probe

Can the candidate effectively track removable pairs using a hash table?
Does the candidate approach the problem using greedy techniques?
Is the candidate able to optimize the algorithm to reduce unnecessary steps?

Common Pitfalls or Variants

Common pitfalls

Failing to consider the order of removals, which could lead to unnecessary operations.
Not efficiently using hash lookups to reduce the number of pair checks.
Misunderstanding the definition of 'non-decreasing' and attempting unnecessary operations.

Follow-up variants

What if the array contains only one element?
Can this problem be solved by applying dynamic programming?
How would the approach change if you could remove more than one pair per operation?

FAQ

What is the minimum number of operations for the array [5,2,3,1]?

The minimum number of operations is 2, as the array becomes non-decreasing after two pair removals.

How do hash lookups help in solving this problem?

Hash lookups allow you to efficiently track which pairs are removable, reducing the number of unnecessary checks.

What if the array is already non-decreasing?

If the array is already non-decreasing, no operations are needed, and the output is 0.

How can I optimize the solution for larger arrays?

Optimizing the solution involves using hash sets for fast lookups and minimizing redundant checks by scanning only the necessary elements.

What is the significance of greedy pair removal in this problem?

Greedy pair removal ensures that the largest possible subset is removed with each operation, minimizing the total number of operations required.

terminal

Solution

Solution 1: Simulation

We define a function $\text{is\_non\_decreasing}(a)$ to determine whether the array $a$ is a non-decreasing array.

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class Solution:
    def minimumPairRemoval(self, nums: List[int]) -> int:
        arr = nums[:]
        ans = 0

        def is_non_decreasing(a: List[int]) -> bool:
            for i in range(1, len(a)):
                if a[i] < a[i - 1]:
                    return False
            return True

        while not is_non_decreasing(arr):
            k = 0
            s = arr[0] + arr[1]
            for i in range(1, len(arr) - 1):
                t = arr[i] + arr[i + 1]
                if s > t:
                    s = t
                    k = i
            arr[k] = s
            arr.pop(k + 1)
            ans += 1
        return ans

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