<p>Given an integer array <code>nums</code>, return <em>the number of all the <strong>arithmetic subsequences</strong> of</em> <code>nums</code>.</p>

<p>A sequence of numbers is called arithmetic if it consists of <strong>at least three elements</strong> and if the difference between any two consecutive elements is the same.</p>

<ul>
	<li>For example, <code>[1, 3, 5, 7, 9]</code>, <code>[7, 7, 7, 7]</code>, and <code>[3, -1, -5, -9]</code> are arithmetic sequences.</li>
	<li>For example, <code>[1, 1, 2, 5, 7]</code> is not an arithmetic sequence.</li>
</ul>

<p>A <strong>subsequence</strong> of an array is a sequence that can be formed by removing some elements (possibly none) of the array.</p>

<ul>
	<li>For example, <code>[2,5,10]</code> is a subsequence of <code>[1,2,1,<strong><u>2</u></strong>,4,1,<u><strong>5</strong></u>,<u><strong>10</strong></u>]</code>.</li>
</ul>

<p>The test cases are generated so that the answer fits in <strong>32-bit</strong> integer.</p>

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

<pre>
<strong>Input:</strong> nums = [2,4,6,8,10]
<strong>Output:</strong> 7
<strong>Explanation:</strong> All arithmetic subsequence slices are:
[2,4,6]
[4,6,8]
[6,8,10]
[2,4,6,8]
[4,6,8,10]
[2,4,6,8,10]
[2,6,10]
</pre>

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

<pre>
<strong>Input:</strong> nums = [7,7,7,7,7]
<strong>Output:</strong> 16
<strong>Explanation:</strong> Any subsequence of this array is arithmetic.
</pre>

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

<ul>
	<li><code>1&nbsp; &lt;= nums.length &lt;= 1000</code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& nums) {
        unordered_map<int, vector<int>> nums2mp;
        int len=nums.size();
        vector<vector<int>> dp(len, vector<int>(len));
        int res=0;
        for(int i=0; i<len; i++)
            nums2mp[nums[i]].push_back(i);
        
        for(int i=0; i<len; i++)
        {
            for(int j=0; j<i; j++){
                long long prev=2ll*nums[j] - nums[i];
                if(prev>INT_MAX || prev<INT_MIN)
                    continue;
                auto item = nums2mp.find(prev);
                if(item!=nums2mp.end())
                {
                    for(auto&k: item->second)
                    {
                        if(k>=j) break;
                        dp[i][j]+=dp[j][k]+1;
                    }
                }
                res+=dp[i][j];
            }
            
        }
        return res;
    }
};
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