2422. Merge Operations to Turn Array Into a Palindrome

Problem Description

You are given an array called nums that contains only positive integers. Your task is to make this array a palindrome. A palindrome is a sequence that reads the same forward and backward. To do this, you are allowed to perform as many operations as you want. An operation consists of selecting any two adjacent elements in nums and replacing them with their sum. The goal is to find the minimum number of such operations required to turn the array into a palindrome.

Intuition

The key to solving this problem lies in understanding how a palindrome is structured: the values on the left side of the array must be mirrored on the right side. The strategy is to iteratively make the sums of the values from both ends (left and right) of the array equal, so they form a palindrome.

To implement this strategy, two pointers approach is used, one starting at the left end (beginning of the array) and one at the right end (end of the array). We compare the sums of the elements at these pointers.

• If the left sum is less than the right sum, this means we need to increase the left sum by moving the left pointer to the right and adding the value at the new pointer position to the left sum, counting this action as one operation.
• Conversely, if the right sum is less than the left sum, we move the right pointer to the left, add the value at the new pointer to the right sum, and count it as an operation as well.
• When both sums are equal, we effectively have a palindrome within those boundaries. We move both pointers inward, skipping over the elements we just confirmed as part of the palindrome because they do not need any more operations.
• This process repeats until the pointers meet, which would mean the entire array has become a palindrome.

The trick here is that we don’t need to actually replace numbers or keep track of the modified array; instead, we just need to know the count of operations required, which is tallied every time we move either pointer to adjust the sums.

The solution's complexity is O(n), where n is the length of the array, because we possibly go through the array only once with our two pointers.

Learn more about Greedy and Two Pointers patterns.

Solution Approach

The solution uses a two-pointers algorithm to walk through the array from both ends towards the center. This approach helps in reducing the problem to smaller subproblems by considering the current sum at both ends. Here's a step-by-step walkthrough:

• Initialize two pointers, i at the start and j at the end of the array.
• Initialize two variables, a and b, to keep track of the sum of the numbers pointed by i and j. Initially, a is assigned nums[i], and b is assigned nums[j].
• Initialize a counter ans with the value 0 to keep track of the number of operations performed.
• Enter a while loop, which will continue to execute as long as i < j (ensuring that we are not comparing the same element with itself or crossing over, which would mean the entire array is a palindrome):
  • Compare the values of a and b.
  • If a < b, we need to increase a to eventually match b. We do so by incrementing i (move the left pointer to the right), adding nums[i] to a, and incrementing the counter ans to represent an operation performed.
  • If b < a, similarly, we need to increase b to match a. Decrement j (move the right pointer to the left), add nums[j] to b, and increment the counter ans.
  • If a == b, it means that the values from nums[i] to nums[j] can be part of the palindrome. Therefore, we increment i, decrement j, and update a and b with the values at the new indices nums[i] and nums[j], respectively. No operation is counted in this case as a and b are already equal.
• Continue the loop until all elements have been accounted for in pairs that form the palindrome.
• The iterator ans is returned as it now contains the minimum number of operations needed to make the array a palindrome.

This simple yet elegant solution leverages the two-pointer technique, which is efficient when you need to compare or pair up elements from opposite ends of an array. It skillfully avoids the need for extra space to store interim arrays, mutating only counters and making the solution very space-efficient (O(1) space complexity).

Example Walkthrough

Let's consider an example to understand the solution approach:

Suppose we are given the following array nums:

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

To make nums a palindrome using the fewest operations, we will follow the steps outlined:

1. Initialize two pointers, i at the start (0) and j at the end (4) of the array.
2. Initialize variables a with nums[i] (which is 1) and b with nums[j] (which is 2).
3. Initialize the operation counter ans to 0.

So, the initial setting looks like:

1i = 0 → [1, 3, 4, 2, 2] ← j = 4
2a = 1
3b = 2
4ans = 0

Now, we start iterating:

• Since a (1) < b (2), we increment i to 1 and update a by adding nums[i], which makes a = 1 + 3 = 4 and ans becomes 1.
• At this point, a (4) == b (2), but for the array to be palindrome, a and b must have the same sum. Thus, we decrease j to 3 and update b by adding nums[j], now b = 2 + 2 = 4, and ans becomes 2.
• Now a (4) == b (4), so we increment i to 2, and decrement j to 2, effectively skipping over the elements we just confirmed to form the correct structure in our palindrome.

Finally, since i now equals j, we've considered the entire array, so we finish.

At the end of the process, ans equals 2, indicating the minimum number of operations required to turn the array into a palindrome.

So, our example array nums can be transformed into a palindrome with a minimum of 2 operations:

• Combine nums[0] and nums[1] to form [4, 4, 2, 2]
• Combine nums[2] and nums[3] to form [4, 6, 2]
• No further operations are needed, as [4, 6, 2] is already a palindrome.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code iterates through the nums list using two pointers i and j that start at opposite ends of the list and move towards the center. The main loop runs while i < j, so in the worst case, it may iterate through all the elements once. Therefore, the worst-case time complexity is O(n), where n is the number of elements in the nums list.

In each iteration of the while loop, the code performs constant-time operations—comparisons and basic arithmetic—so these do not affect the overall O(n) time complexity.

Space Complexity

Since the algorithm operates in place and the amount of additional memory used does not depend on the input size (nums list), the space complexity is O(1). Only a fixed number of variables i, j, a, b, and ans are used, which occupies constant space irrespective of the input size.

Learn more about how to find time and space complexity quickly using problem constraints.