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Minimum Pair Removal to Sort Array I

Easy

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Topics:

ArraysGreedy AlgorithmsStacks

Given an array nums, you can perform the following operation any number of times:

Select the adjacent pair with the minimum sum in nums. If multiple such pairs exist, choose the leftmost one.
Replace the pair with their sum.

Return the minimum number of operations needed to make the array non-decreasing.

An array is said to be non-decreasing if each element is greater than or equal to its previous element (if it exists).

Example 1:

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

Output: 2

Explanation:

The pair (3,1) has the minimum sum of 4. After replacement, nums = [5,2,4].
The pair (2,4) has the minimum sum of 6. After replacement, nums = [5,6].

The array nums became non-decreasing in two operations.

Example 2:

Input: nums = [1,2,2]

Output: 0

Explanation:

The array nums is already sorted.

Constraints:

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

Try coding it in LeetCode

Minimum Pair Removal to Sort Array I

Easy

Asked by:

34 views

Topics:

ArraysGreedy AlgorithmsStacks

Given an array nums, you can perform the following operation any number of times:

Select the adjacent pair with the minimum sum in nums. If multiple such pairs exist, choose the leftmost one.
Replace the pair with their sum.

Return the minimum number of operations needed to make the array non-decreasing.

An array is said to be non-decreasing if each element is greater than or equal to its previous element (if it exists).

Example 1:

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

Output: 2

Explanation:

The pair (3,1) has the minimum sum of 4. After replacement, nums = [5,2,4].
The pair (2,4) has the minimum sum of 6. After replacement, nums = [5,6].

The array nums became non-decreasing in two operations.

Example 2:

Input: nums = [1,2,2]

Output: 0

Explanation:

The array nums is already sorted.

Constraints:

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of integer values in the input array? Can they be negative?
Can the input array be empty, or contain only one element?
Are duplicate values allowed in the array, and if so, how should they be handled?
By 'sorted', do you mean non-decreasing order (i.e., ascending with possible duplicates) or strictly increasing order?
If there are multiple possible sets of minimum pairs to remove, is any valid set acceptable?

Brute Force Solution

Approach

The basic idea is to try every single possible combination of removing pairs from the input. We then check if the remaining items are in order. Finally, we pick the combination that removed the fewest pairs while resulting in a sorted collection of items.

Here's how the algorithm would work step-by-step:

Consider all possible ways to pick pairs of items to remove from the collection.
For each selection of pairs to remove, create a new collection with those pairs removed.
Check if the new collection is sorted from smallest to largest.
If the new collection is sorted, record the number of pairs that were removed.
After checking all possible removals, find the smallest number of pairs that needed to be removed to achieve a sorted collection.
Report that smallest number.

Code Implementation

def minimum_pair_removal_to_sort_array_brute_force(numbers):
    number_of_elements = len(numbers)
    minimum_pairs_removed = number_of_elements

    for i in range(1 << number_of_elements):
        elements_to_keep = []
        pairs_removed = 0

        # Iterate through the input array and choose elements to keep
        for j in range(number_of_elements):
            if (i >> j) & 1:
                elements_to_keep.append(numbers[j])

        # Calculate the number of pairs that were removed
        pairs_removed = (number_of_elements - len(elements_to_keep)) // 2

        # Check if the remaining elements are sorted
        is_sorted = True
        for k in range(len(elements_to_keep) - 1):
            if elements_to_keep[k] > elements_to_keep[k + 1]:
                is_sorted = False
                break

        # If the array is sorted, check if it's the minimum
        if is_sorted:
            # Here, we found a sorted array, so we check if
            # the number of removed pairs is a new minimum.
            minimum_pairs_removed = min(minimum_pairs_removed, pairs_removed)

    return minimum_pairs_removed

Big(O) Analysis

Time Complexity

O(2^n * n) – The algorithm explores all possible combinations of removing pairs. In the worst case, we can have n/2 pairs, so we're looking at 2^(n/2) or 2^n possible combinations to check. For each combination, we need to create a new collection which takes O(n) time in the worst case, then check if the collection is sorted, which again takes O(n) time. We multiply the number of combinations by the cost of creating and checking the collection. This results in a time complexity of approximately O(2^n * n).

Space Complexity

O(2^N) – The algorithm considers all possible combinations of removing pairs, which translates to generating many sub-collections. In the worst case, the number of sub-collections checked grows exponentially. The space required to store each sub-collection in step 2 can be at most N, where N is the number of items in the original collection, but the number of these sub-collections is O(2^(N/2)) since we consider pairs, leading to an overall space complexity dominated by the number of combinations generated. Therefore, the space complexity is O(2^N).

Optimal Solution

Approach

The core idea is to efficiently count how many elements we can keep without creating disorder. We focus on preserving the longest possible chain of elements that are already in order, rather than exhaustively checking every possible removal.

Here's how the algorithm would work step-by-step:

Start by looking at the given set of numbers from left to right.
Keep track of the length of the longest increasing sequence you've seen so far.
As you go through the numbers, if you find a number that is greater than or equal to the last number you kept, add it to your sequence.
If you find a number that is smaller than the last number you kept, you will need to check if it can replace an element in your current sequence to make it better.
If the current number can replace an element in the sequence so that it remains an increasing sequence, then replace it.
At the end, the number of removals will be the original number of elements minus the length of the longest increasing sequence.

Code Implementation

def minimum_pair_removal_to_sort_array(numbers):
    longest_increasing_sequence = []

    for number in numbers:
        # If the current number extends the sequence.
        if not longest_increasing_sequence or number >= longest_increasing_sequence[-1]:
            longest_increasing_sequence.append(number)

        else:
            # Find the smallest number in the sequence that is >= current.
            left_index = 0
            right_index = len(longest_increasing_sequence) - 1

            while left_index <= right_index:
                middle_index = (left_index + right_index) // 2

                if longest_increasing_sequence[middle_index] <= number:
                    left_index = middle_index + 1
                else:
                    right_index = middle_index - 1

            # Replace the found element with the current number.
            # This keeps the sequence increasing and potentially shorter.
            if left_index < len(longest_increasing_sequence):
                longest_increasing_sequence[left_index] = number

    # Remaining elements need to be removed.
    number_of_removals = len(numbers) - len(longest_increasing_sequence)
    return number_of_removals

Big(O) Analysis

Time Complexity

O(n log n) – We iterate through the array of n elements once. Inside the loop, if an element is smaller than the last element of our increasing sequence, we perform a binary search (log n) to find the correct position to replace an element in the increasing sequence (which maintains the longest increasing subsequence seen so far). Thus, the overall time complexity is O(n log n) due to the n iterations with a log n binary search in each.

Space Complexity

O(N) – The algorithm maintains a sequence to track the longest increasing subsequence. In the worst-case scenario where the input array is strictly decreasing, every element could potentially replace an element in the sequence, leading to the sequence growing up to the size of the input array. Thus, the auxiliary space used for storing this sequence is proportional to the input size, N. Therefore, the space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as no pairs can be formed.
Input array with a single element	Return 0 since a pair requires at least two elements.
Input array already sorted	Return 0 as no elements need to be removed.
Input array in reverse sorted order	The algorithm should correctly identify and remove the necessary pairs to achieve sorting.
Array with all elements identical	Return array size minus 1 as only one unique element would fulfill the requirement for a sorted array.
Large array size exceeding memory limits	The algorithm should use an efficient in-place approach or be mindful of memory usage to avoid out-of-memory errors.
Array containing negative numbers, zeros, and positive numbers	The sorting logic must correctly handle the full range of integer values.
Array where all possible pairs need to be removed	The algorithm must iterate through all necessary pairs to reach a sorted array.

Null or empty input array

How to Handle:

Return 0 immediately as no pairs can be formed.

Input array with a single element

How to Handle:

Return 0 since a pair requires at least two elements.

Input array already sorted

How to Handle:

Return 0 as no elements need to be removed.

Input array in reverse sorted order

How to Handle:

The algorithm should correctly identify and remove the necessary pairs to achieve sorting.

Array with all elements identical

How to Handle:

Return array size minus 1 as only one unique element would fulfill the requirement for a sorted array.

Large array size exceeding memory limits

How to Handle:

The algorithm should use an efficient in-place approach or be mindful of memory usage to avoid out-of-memory errors.

Array containing negative numbers, zeros, and positive numbers

How to Handle:

The sorting logic must correctly handle the full range of integer values.

Array where all possible pairs need to be removed

How to Handle:

The algorithm must iterate through all necessary pairs to reach a sorted array.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of integer values in the input array? Can they be negative?
Can the input array be empty, or contain only one element?
Are duplicate values allowed in the array, and if so, how should they be handled?
By 'sorted', do you mean non-decreasing order (i.e., ascending with possible duplicates) or strictly increasing order?
If there are multiple possible sets of minimum pairs to remove, is any valid set acceptable?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Consider all possible ways to pick pairs of items to remove from the collection.
For each selection of pairs to remove, create a new collection with those pairs removed.
Check if the new collection is sorted from smallest to largest.
If the new collection is sorted, record the number of pairs that were removed.
After checking all possible removals, find the smallest number of pairs that needed to be removed to achieve a sorted collection.
Report that smallest number.

Code Implementation

def minimum_pair_removal_to_sort_array_brute_force(numbers):
    number_of_elements = len(numbers)
    minimum_pairs_removed = number_of_elements

    for i in range(1 << number_of_elements):
        elements_to_keep = []
        pairs_removed = 0

        # Iterate through the input array and choose elements to keep
        for j in range(number_of_elements):
            if (i >> j) & 1:
                elements_to_keep.append(numbers[j])

        # Calculate the number of pairs that were removed
        pairs_removed = (number_of_elements - len(elements_to_keep)) // 2

        # Check if the remaining elements are sorted
        is_sorted = True
        for k in range(len(elements_to_keep) - 1):
            if elements_to_keep[k] > elements_to_keep[k + 1]:
                is_sorted = False
                break

        # If the array is sorted, check if it's the minimum
        if is_sorted:
            # Here, we found a sorted array, so we check if
            # the number of removed pairs is a new minimum.
            minimum_pairs_removed = min(minimum_pairs_removed, pairs_removed)

    return minimum_pairs_removed

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start by looking at the given set of numbers from left to right.
Keep track of the length of the longest increasing sequence you've seen so far.
As you go through the numbers, if you find a number that is greater than or equal to the last number you kept, add it to your sequence.
If you find a number that is smaller than the last number you kept, you will need to check if it can replace an element in your current sequence to make it better.
If the current number can replace an element in the sequence so that it remains an increasing sequence, then replace it.
At the end, the number of removals will be the original number of elements minus the length of the longest increasing sequence.

Code Implementation

def minimum_pair_removal_to_sort_array(numbers):
    longest_increasing_sequence = []

    for number in numbers:
        # If the current number extends the sequence.
        if not longest_increasing_sequence or number >= longest_increasing_sequence[-1]:
            longest_increasing_sequence.append(number)

        else:
            # Find the smallest number in the sequence that is >= current.
            left_index = 0
            right_index = len(longest_increasing_sequence) - 1

            while left_index <= right_index:
                middle_index = (left_index + right_index) // 2

                if longest_increasing_sequence[middle_index] <= number:
                    left_index = middle_index + 1
                else:
                    right_index = middle_index - 1

            # Replace the found element with the current number.
            # This keeps the sequence increasing and potentially shorter.
            if left_index < len(longest_increasing_sequence):
                longest_increasing_sequence[left_index] = number

    # Remaining elements need to be removed.
    number_of_removals = len(numbers) - len(longest_increasing_sequence)
    return number_of_removals

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as no pairs can be formed.
Input array with a single element	Return 0 since a pair requires at least two elements.
Input array already sorted	Return 0 as no elements need to be removed.
Input array in reverse sorted order	The algorithm should correctly identify and remove the necessary pairs to achieve sorting.
Array with all elements identical	Return array size minus 1 as only one unique element would fulfill the requirement for a sorted array.
Large array size exceeding memory limits	The algorithm should use an efficient in-place approach or be mindful of memory usage to avoid out-of-memory errors.
Array containing negative numbers, zeros, and positive numbers	The sorting logic must correctly handle the full range of integer values.
Array where all possible pairs need to be removed	The algorithm must iterate through all necessary pairs to reach a sorted array.

Null or empty input array

How to Handle:

Return 0 immediately as no pairs can be formed.

Input array with a single element

How to Handle:

Return 0 since a pair requires at least two elements.

Input array already sorted

How to Handle:

Return 0 as no elements need to be removed.

Input array in reverse sorted order

How to Handle:

The algorithm should correctly identify and remove the necessary pairs to achieve sorting.

Array with all elements identical

How to Handle:

Return array size minus 1 as only one unique element would fulfill the requirement for a sorted array.

Large array size exceeding memory limits

How to Handle:

The algorithm should use an efficient in-place approach or be mindful of memory usage to avoid out-of-memory errors.

Array containing negative numbers, zeros, and positive numbers

How to Handle:

The sorting logic must correctly handle the full range of integer values.

Array where all possible pairs need to be removed

How to Handle:

The algorithm must iterate through all necessary pairs to reach a sorted array.