#2856

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Minimum Array Length After Pair Removals

This problem involves minimizing the length of a sorted array by repeatedly removing adjacent pairs of equal elements.

array hash table two pointers binary search greedy

Problem Statement

You are given an integer array nums sorted in non-decreasing order. Your task is to repeatedly remove adjacent pairs of equal elements, minimizing the array length.

You can perform the operation any number of times. Each operation removes a pair of adjacent equal elements, and the goal is to determine the minimum length of the array after performing zero or more operations.

Examples

Example 1

Input: nums = [1,2,3,4]

Output: 0

Example 2

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

Output: 0

Example 3

Input: nums = [1000000000,1000000000]

Output: 2

Since both numbers are equal, they cannot be removed.

Constraints

1 <= nums.length <= 105
1 <= nums[i] <= 109
nums is sorted in non-decreasing order.

Solution Approach

Array Scanning with Hash Table

We scan through the array while using a hash table to keep track of the frequency of each element. By detecting pairs efficiently, we can remove them during each pass through the array. This approach ensures that we remove as many pairs as possible with each operation.

Two Pointers Technique

Using two pointers, one for the beginning of the array and another for the end, we can compare elements and remove pairs by shifting the pointers accordingly. This reduces unnecessary checks and minimizes the array length faster.

Greedy Approach with Frequency Count

A greedy strategy can be applied by focusing on removing pairs with the highest frequency first, reducing the array length optimally. The frequency count can be handled with a hash table, and the pair removal is based on this count.

Complexity Analysis

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

The time complexity can vary depending on the approach. The array scanning with hash table approach can be done in O(n) time. Space complexity also depends on the method used, but typically it requires O(n) space due to the hash table or auxiliary storage used for element counts.

What Interviewers Usually Probe

Look for efficiency in handling large input arrays (up to 10^5 elements).
Evaluate the candidate's approach to minimizing operations and array length.
Assess their understanding of array scanning and hash lookups in reducing problem complexity.

Common Pitfalls or Variants

Common pitfalls

Not considering the sorted nature of the array could lead to unnecessary comparisons.
Failing to efficiently track element frequencies, causing an excessive number of operations.
Overcomplicating the solution by not using hash tables or efficient frequency counting methods.

Follow-up variants

Consider unsorted arrays or arrays with more varied element frequencies.
Explore variations with constraints on the number of allowed operations.
What if the array contains only one element or no removable pairs?

FAQ

What is the main pattern used to solve the Minimum Array Length After Pair Removals problem?

The main pattern is array scanning combined with hash lookup. Efficient frequency counting allows us to identify and remove adjacent pairs.

How can two pointers help solve the problem?

Two pointers allow for efficient comparison of elements from both ends, removing adjacent pairs and minimizing the array length faster.

What is the time complexity of the solution for the Minimum Array Length After Pair Removals problem?

The time complexity can be O(n) depending on the solution approach, particularly when using array scanning with a hash table or two pointers.

Can the problem be solved without using a hash table?

While a hash table is efficient, alternative methods like two-pointer techniques can also be used to minimize operations and remove pairs.

How does GhostInterview help with this problem?

GhostInterview helps by guiding you through efficient approaches and avoiding common pitfalls, ensuring you maximize the number of operations for an optimal solution.

terminal

Solution

Solution 1: Greedy + Priority Queue (Max Heap)

We use a hash table $cnt$ to count the occurrence of each element in the array $nums$, then add each value in $cnt$ to a priority queue (max heap) $pq$. Each time we take out two elements $x$ and $y$ from $pq$, decrease their values by one. If the value after decrement is still greater than $0$, we add the decremented value back to $pq$. Each time we take out two elements from $pq$, it means we delete a pair of numbers from the array, so the length of the array decreases by $2$. When the size of $pq$ is less than $2$, we stop the deletion operation.

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class Solution:
    def minLengthAfterRemovals(self, nums: List[int]) -> int:
        cnt = Counter(nums)
        pq = [-x for x in cnt.values()]
        heapify(pq)
        ans = len(nums)
        while len(pq) > 1:
            x, y = -heappop(pq), -heappop(pq)
            x -= 1
            y -= 1
            if x > 0:
                heappush(pq, -x)
            if y > 0:
                heappush(pq, -y)
            ans -= 2
        return ans

Continue Practicing

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