2856. Minimum Array Length After Pair Removals

LeetCode problem 2856

Problem Statement

Given a 0-indexed sorted array of integers nums, you can perform a specific operation an unlimited number of times:

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

The task is to determine and return the smallest possible length of nums after executing the operation as many times as you wish.

Naive Solution

One way to approach this problem would be to iterate through every possible pair of elements in nums to check if they satisfy the condition nums[i] < nums[j]. Whenever a valid pair is found, remove them and restart the search. This method would be inefficient and would result in a high time complexity due to frequent list updates.

Hints & Tips

• Utilizing two pointers can help in efficiently determining the pairs to remove.
• Keep track of removed indices in a set to avoid duplication.
• Focus on removing the largest numbers since they have the most potential pairs.

Approach

The idea is to maintain two pointers: a slow pointer i starting from the beginning of the array and a fast pointer j starting from the middle of the array. Since the array is sorted, this ensures that the number at j is always greater than the number at i.

1. Traverse the list with the two pointers.
2. If a valid pair is found (i.e., nums[i] < nums[j] and i hasn’t been removed yet), mark the indices i and j as removed.
3. Move the pointer i one step forward.
4. Repeat the process until the end of the array is reached.

The result would be the initial length of nums subtracted by the number of removed indices.

Steps

1. Initialize two pointers: i = 0 and j = len(nums) // 2.
2. Create a set removed to keep track of removed indices.
3. Traverse the list with the two pointers.
4. If nums[j] > nums[i] and i hasn’t been removed, add i and j to the removed set and move the pointer i one step forward.
5. Continue the process until the end of the array.
6. The result is len(nums) - len(removed).

Solution

def minimumLengthAfterRemoval(nums):
    i = 0
    removed = set()
    for j in range(len(nums) // 2, len(nums)):
        if nums[j] > nums[i] and i not in removed:
            removed.add(i)
            removed.add(j)
            i += 1
    return len(nums) - len(removed)