# 2422. Merge Operations to Turn Array Into a Palindrome

## Description

You are given an array nums consisting of positive integers.

You can perform the following operation on the array any number of times:

• Choose any two adjacent elements and replace them with their sum.
• For example, if nums = [1,2,3,1], you can apply one operation to make it [1,5,1].

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

Example 1:

Input: nums = [4,3,2,1,2,3,1]
Output: 2
Explanation: We can turn the array into a palindrome in 2 operations as follows:
- Apply the operation on the fourth and fifth element of the array, nums becomes equal to [4,3,2,3,3,1].
- Apply the operation on the fifth and sixth element of the array, nums becomes equal to [4,3,2,3,4].
The array [4,3,2,3,4] is a palindrome.
It can be shown that 2 is the minimum number of operations needed.


Example 2:

Input: nums = [1,2,3,4]
Output: 3
Explanation: We do the operation 3 times in any position, we obtain the array [10] at the end which is a palindrome.


Constraints:

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

## Solutions

Solution 1: Greedy + Two Pointers

Define two pointers $i$ and $j$, pointing to the beginning and end of the array respectively, use variables $a$ and $b$ to represent the values of the first and last elements, and variable $ans$ to represent the number of operations.

If $a < b$, we move the pointer $i$ one step to the right, i.e., $i \leftarrow i + 1$, then add the value of the element pointed to by $i$ to $a$, i.e., $a \leftarrow a + nums[i]$, and increment the operation count by one, i.e., $ans \leftarrow ans + 1$.

If $a > b$, we move the pointer $j$ one step to the left, i.e., $j \leftarrow j - 1$, then add the value of the element pointed to by $j$ to $b$, i.e., $b \leftarrow b + nums[j]$, and increment the operation count by one, i.e., $ans \leftarrow ans + 1$.

Otherwise, it means $a = b$, at this time we move the pointer $i$ one step to the right, i.e., $i \leftarrow i + 1$, move the pointer $j$ one step to the left, i.e., $j \leftarrow j - 1$, and update the values of $a$ and $b$, i.e., $a \leftarrow nums[i]$ and $b \leftarrow nums[j]$.

Repeat the above process until $i \ge j$, return the operation count $ans$.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

• class Solution {
public int minimumOperations(int[] nums) {
int i = 0, j = nums.length - 1;
long a = nums[i], b = nums[j];
int ans = 0;
while (i < j) {
if (a < b) {
a += nums[++i];
++ans;
} else if (b < a) {
b += nums[--j];
++ans;
} else {
a = nums[++i];
b = nums[--j];
}
}
return ans;
}
}

• class Solution {
public:
int minimumOperations(vector<int>& nums) {
int i = 0, j = nums.size() - 1;
long a = nums[i], b = nums[j];
int ans = 0;
while (i < j) {
if (a < b) {
a += nums[++i];
++ans;
} else if (b < a) {
b += nums[--j];
++ans;
} else {
a = nums[++i];
b = nums[--j];
}
}
return ans;
}
};

• class Solution:
def minimumOperations(self, nums: List[int]) -> int:
i, j = 0, len(nums) - 1
a, b = nums[i], nums[j]
ans = 0
while i < j:
if a < b:
i += 1
a += nums[i]
ans += 1
elif b < a:
j -= 1
b += nums[j]
ans += 1
else:
i, j = i + 1, j - 1
a, b = nums[i], nums[j]
return ans


• func minimumOperations(nums []int) int {
i, j := 0, len(nums)-1
a, b := nums[i], nums[j]
ans := 0
for i < j {
if a < b {
i++
a += nums[i]
ans++
} else if b < a {
j--
b += nums[j]
ans++
} else {
i, j = i+1, j-1
a, b = nums[i], nums[j]
}
}
return ans
}