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
class Solution {
    private Integer[][] f;
    private int[] nums;
    private int s;
    private int n;

    public int maxOperations(int[] nums) {
        this.nums = nums;
        n = nums.length;
        int a = g(2, n - 1, nums[0] + nums[1]);
        int b = g(0, n - 3, nums[n - 2] + nums[n - 1]);
        int c = g(1, n - 2, nums[0] + nums[n - 1]);
        return 1 + Math.max(a, Math.max(b, c));
    }

    private int g(int i, int j, int s) {
        f = new Integer[n][n];
        this.s = s;
        return dfs(i, j);
    }

    private int dfs(int i, int j) {
        if (j - i < 1) {
            return 0;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        int ans = 0;
        if (nums[i] + nums[i + 1] == s) {
            ans = Math.max(ans, 1 + dfs(i + 2, j));
        }
        if (nums[i] + nums[j] == s) {
            ans = Math.max(ans, 1 + dfs(i + 1, j - 1));
        }
        if (nums[j - 1] + nums[j] == s) {
            ans = Math.max(ans, 1 + dfs(i, j - 2));
        }
        return f[i][j] = ans;
    }
}