Merge Operations to Turn Array Into a Palindrome

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ArraysTwo PointersGreedy Algorithms

You are given an array nums consisting only of positive integers.

You are allowed to perform the following operation on the array any number of times:

Choose any two adjacent elements of the array and replace them with their sum.

For example, if nums = [1, 2, 3, 1], you can apply one operation to make it nums = [3, 3, 1].

Return the minimum number of operations needed to make the array a palindrome.

Note: An array is a palindrome if nums[i] == nums[n - i - 1] for all 0 <= i < n/2.

Example 1:

Input: nums = [4,3,2,1,2,3,1]
Output: 2
Explanation: We can perform the following operations:
- Add 4 and 3 to make nums = [7,2,1,2,3,1].
- Add 1 and 3 to make nums = [7,2,1,2,4].
The array is now a palindrome [7,2,1,2,7].
It can be shown that 2 is the minimum number of operations needed.

Example 2:

Input: nums = [1,2,3,4]
Output: 3
Explanation: There is no way to make the array a palindrome with less than 3 operations.

Example 3:

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

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁶

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 that can be present in the array?
Can the input array be empty or null?
If the array cannot be transformed into a palindrome through merges, what should I return?
Are we only allowed to merge adjacent elements, and does the order of merging matter?
If there are multiple ways to achieve a palindrome with the same number of merge operations, is any one solution acceptable?

Brute Force Solution

Approach

The brute force method for making an array a palindrome involves exploring every single possible merge combination. We try merging different pairs of numbers and after each merge, we check if the resulting array is a palindrome. We continue this process until we either find a palindrome or have exhausted all combinations.

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

Start with the original array.
Consider every possible pair of numbers that can be merged. This includes merging the first two numbers, merging the last two numbers, and possibly other combinations depending on the size of the array.
For each possible merge, create a new array with the merged numbers.
Check if the new array is a palindrome. A palindrome reads the same forwards and backward.
If the new array is a palindrome, you've found a solution! Record the number of merges performed.
If the new array is not a palindrome, repeat the process of merging pairs of numbers within the new array, creating even more arrays.
Continue this process of merging and checking for palindromes until you have explored all possible merging combinations.
If you find multiple palindromes, choose the one that required the fewest merges. If no palindromes are found after trying all merges, then there is no solution using this method.

Code Implementation

def merge_operations_to_palindrome_brute_force(original_array):
    minimum_merges = float('inf')
    queue = [(original_array, 0)]

    while queue:
        current_array, current_merges = queue.pop(0)

        if current_array == current_array[::-1]:
            minimum_merges = min(minimum_merges, current_merges)
            continue

        # Only explore further if the current number of merges is less than the minimum found so far
        if current_merges >= minimum_merges:
            continue

        for index in range(len(current_array) - 1):

            # Create a new array by merging two adjacent elements
            new_array = current_array[:index] + [current_array[index] + current_array[index + 1]] + current_array[index + 2:]

            queue.append((new_array, current_merges + 1))

    if minimum_merges == float('inf'):
        return -1
    else:
        return minimum_merges

Big(O) Analysis

Time Complexity

O(2^n) – The described brute force approach explores all possible merge combinations. At each step, we consider merging pairs, which effectively creates a branching tree of possibilities. In the worst-case scenario, we might explore a significant portion of all possible subsets of the array to determine merge candidates. This exponential exploration of merging possibilities results in a time complexity of O(2^n), where n is the number of elements in the input array, as the number of possible combinations grows exponentially with the size of the array. The palindrome check itself takes O(n) time, but the dominant factor is the number of combinations explored.

Space Complexity

O(N!) – The brute force method explores all possible merge combinations through recursive calls. Each recursive call creates a new array, potentially leading to many array copies. In the worst case, the number of possible combinations can grow factorially with the input size N, where N is the length of the original array. Therefore, the space complexity is approximately O(N!) due to the large number of arrays created during the exploration of all merge combinations.

Optimal Solution

Approach

To make the array a palindrome most efficiently, we work from the outside in, comparing the elements at the beginning and end. If they are unequal, we merge the smaller element with its neighbor to try and make the array palindromic.

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

Compare the first and last numbers in the array.
If the numbers are the same, move inward to the next pair.
If the first number is smaller than the last number, add it to the number next to it, and treat this sum as the new first number. Essentially, we are merging the first two numbers.
If the last number is smaller than the first number, add it to the number before it, and treat this sum as the new last number. We are merging the last two numbers.
Repeat this process, always moving inward, until only one number is left, or you have determined that the array can be made into a palindrome.
Count how many merge operations you had to perform.

Code Implementation

def merge_operations_to_palindrome(array):
    start_index = 0
    end_index = len(array) - 1
    merge_count = 0

    while start_index < end_index:
        if array[start_index] == array[end_index]:
            start_index += 1
            end_index -= 1

        elif array[start_index] < array[end_index]:
            # Merge from the start to increase value
            array[start_index + 1] += array[start_index]

            start_index += 1
            merge_count += 1

        else:
            # Merge from the end to increase value
            array[end_index - 1] += array[end_index]

            end_index -= 1

            merge_count += 1

    return merge_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the array, comparing the first and last elements. In each step, either the left or the right pointer moves inward. Each element in the array is visited and potentially merged at most once. Therefore, the number of operations is proportional to the number of elements n in the array, leading to a time complexity of O(n).

Space Complexity

O(1) – The algorithm operates in-place, modifying the input array directly. It only uses a few constant space variables, such as the indices of the start and end of the array and a counter for merge operations. Thus, the amount of auxiliary space used does not depend on the size of the input array, N. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Null or undefined input array	Throw an IllegalArgumentException or return an empty array to avoid NullPointerException.
Empty array	Return 0 merge operations as an empty array is already a palindrome.
Array with a single element	Return 0 merge operations as a single-element array is already a palindrome.
Array with two elements that are equal	Return 0 merge operations as no merges are needed.
Array with two elements that are not equal	Return 1 merge operation (merge either end to the other element).
Array with all elements equal	Return 0 merge operations as the array is already a palindrome.
Array with elements that lead to integer overflow during merging	Use a larger data type (e.g., long) or check for potential overflow before merging.
Array that cannot be transformed into a palindrome (e.g., [1,2,3])	The algorithm should proceed until the left and right pointers meet, effectively performing all possible merges to minimize operations even if a true palindrome is not achieved.

Null or undefined input array

How to Handle:

Throw an IllegalArgumentException or return an empty array to avoid NullPointerException.

Empty array

How to Handle:

Return 0 merge operations as an empty array is already a palindrome.

Array with a single element

How to Handle:

Return 0 merge operations as a single-element array is already a palindrome.

Array with two elements that are equal

How to Handle:

Return 0 merge operations as no merges are needed.

Array with two elements that are not equal

How to Handle:

Return 1 merge operation (merge either end to the other element).

Array with all elements equal

How to Handle:

Return 0 merge operations as the array is already a palindrome.

Array with elements that lead to integer overflow during merging

How to Handle:

Use a larger data type (e.g., long) or check for potential overflow before merging.

Array that cannot be transformed into a palindrome (e.g., [1,2,3])

How to Handle:

The algorithm should proceed until the left and right pointers meet, effectively performing all possible merges to minimize operations even if a true palindrome is not achieved.