Is Minimum Array Length After Pair Removals easy or hard?

The problem is rated Medium because the greedy insight is not immediately obvious. Many developers initially try simulation or brute force before recognizing that pairing the smallest available element with the earliest valid larger element guarantees the optimal result.

Minimum Array Length After Pair Removals Python/Java solution

Most implementations use a two-pointer loop that scans the array once. The same logic works across Python, Java, C++, C#, and JavaScript because it only requires pointer movement and simple comparisons, producing an O(n) time solution.

How to solve Minimum Array Length After Pair Removals in O(n)?

Use two pointers on the sorted array. Set i at the start and j at the midpoint. When nums[i] < nums[j], form a pair and move both pointers forward; otherwise move j forward. Count valid pairs and compute the remaining length as n - 2 * pairs.

What is the best approach for Minimum Array Length After Pair Removals?

The most efficient approach uses a greedy two-pointer strategy. Since the array is non-decreasing, start one pointer at the beginning and another near the middle. Pair the smallest element with the earliest larger element that satisfies nums[i] < nums[j]. This produces the maximum number of removable pairs in O(n) time and O(1) space.

Is Minimum Array Length After Pair Removals asked at Google/Amazon/Meta?

Greedy pairing and two-pointer problems like this commonly appear in interviews at companies such as Google, Amazon, and Meta. The problem tests understanding of greedy matching, sorted array traversal, and reasoning about element frequency constraints.

What data structure is used in Minimum Array Length After Pair Removals?

The main techniques rely on arrays and the two-pointer pattern. An alternative solution uses a hash map or counter to track element frequencies and reason about the most frequent value limiting the number of valid pair removals.

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

The optimal solution runs in O(n) time because each element is processed at most once using two pointers. The space complexity is O(1) when using the pointer-based greedy approach. A frequency-count implementation also runs in O(n) time but uses O(n) additional space.

Description Hints 6 Solution Video 17:57

2856. Minimum Array Length After Pair Removals

Q: What is the best approach for Minimum Array Length After Pair Removals?

The most efficient approach uses a greedy two-pointer strategy. Since the array is non-decreasing, start one pointer at the beginning and another near the middle. Pair the smallest element with the earliest larger element that satisfies nums[i] < nums[j]. This produces the maximum number of removable pairs in O(n) time and O(1) space.

Q: Is Minimum Array Length After Pair Removals asked at Google/Amazon/Meta?

Greedy pairing and two-pointer problems like this commonly appear in interviews at companies such as Google, Amazon, and Meta. The problem tests understanding of greedy matching, sorted array traversal, and reasoning about element frequency constraints.

Q: What data structure is used in Minimum Array Length After Pair Removals?

The main techniques rely on arrays and the two-pointer pattern. An alternative solution uses a hash map or counter to track element frequencies and reason about the most frequent value limiting the number of valid pair removals.

Q: What is the time complexity of Minimum Array Length After Pair Removals?

The optimal solution runs in O(n) time because each element is processed at most once using two pointers. The space complexity is O(1) when using the pointer-based greedy approach. A frequency-count implementation also runs in O(n) time but uses O(n) additional space.

Medium27.8% acceptance1 company

Given an integer array num sorted in non-decreasing order.

You can perform the following operation any number of times:

Choose two indices, i and j, where nums[i] < nums[j].
Then, remove the elements at indices i and j from nums. The remaining elements retain their original order, and the array is re-indexed.

Return the minimum length of nums after applying the operation zero or more times.

Example 1:

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

Output: 0

Explanation:

Example 2:

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

Output: 0

Explanation:

Example 3:

Input: nums = [1000000000,1000000000]

Output: 2

Explanation:

Since both numbers are equal, they cannot be removed.

Example 4:

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

Output: 1

Explanation:

Constraints:

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

Asked at 1 company

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Hint 1

To minimize the length of the array, we should maximize the number of operations performed.

Hint 2

To perform k operations, it is optimal to use the smallest k values and the largest k values in nums.

Hint 3

What is the best way to make pairs from the smallest k values and the largest k values so it is possible to remove all the pairs?

Hint 4

If we consider the smallest k values and the largest k values as two separate sorted 0-indexed arrays, a and b, It is optimal to pair a[i] and b[i]. So, a k is valid if a[i] < b[i] for all i in the range [0, k - 1].

Hint 5

The greatest possible valid k can be found using binary search.

Hint 6

The answer is nums.length - 2 * k.

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2856. Minimum Array Length After Pair Removals

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