Maximum Number of Operations With the Same Score II
<p>Given an array of integers called <code>nums</code>, you can perform <strong>any</strong> of the following operation while <code>nums</code> contains <strong>at least</strong> <code>2</code> elements:</p>

<ul>
	<li>Choose the first two elements of <code>nums</code> and delete them.</li>
	<li>Choose the last two elements of <code>nums</code> and delete them.</li>
	<li>Choose the first and the last elements of <code>nums</code> and delete them.</li>
</ul>

<p>The<strong> score</strong> of the operation is the sum of the deleted elements.</p>

<p>Your task is to find the <strong>maximum</strong> number of operations that can be performed, such that <strong>all operations have the same score</strong>.</p>

<p>Return <em>the <strong>maximum</strong> number of operations possible that satisfy the condition mentioned above</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,2,1,2,3,4]
<strong>Output:</strong> 3
<strong>Explanation:</strong> We perform the following operations:
- Delete the first two elements, with score 3 + 2 = 5, nums = [1,2,3,4].
- Delete the first and the last elements, with score 1 + 4 = 5, nums = [2,3].
- Delete the first and the last elements, with score 2 + 3 = 5, nums = [].
We are unable to perform any more operations as nums is empty.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,2,6,1,4]
<strong>Output:</strong> 2
<strong>Explanation:</strong> We perform the following operations:
- Delete the first two elements, with score 3 + 2 = 5, nums = [6,1,4].
- Delete the last two elements, with score 1 + 4 = 5, nums = [6].
It can be proven that we can perform at most 2 operations.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums.length &lt;= 2000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
</ul>
Memoization; Array; Dynamic Programming
func maxOperations(nums []int) int {
	n := len(nums)
	var g func(i, j, s int) int
	g = func(i, j, s int) int {
		f := make([][]int, n)
		for i := range f {
			f[i] = make([]int, n)
			for j := range f {
				f[i][j] = -1
			}
		}
		var dfs func(i, j int) int
		dfs = func(i, j int) int {
			if j-i < 1 {
				return 0
			}
			if f[i][j] != -1 {
				return f[i][j]
			}
			ans := 0
			if nums[i]+nums[i+1] == s {
				ans = max(ans, 1+dfs(i+2, j))
			}

			if nums[i]+nums[j] == s {
				ans = max(ans, 1+dfs(i+1, j-1))
			}

			if nums[j-1]+nums[j] == s {
				ans = max(ans, 1+dfs(i, j-2))
			}
			f[i][j] = ans
			return ans
		}
		return dfs(i, j)
	}
	a := g(2, n-1, nums[0]+nums[1])
	b := g(0, n-3, nums[n-1]+nums[n-2])
	c := g(1, n-2, nums[0]+nums[n-1])
	return 1 + max(a, b, c)
}