<p>Given an integer array <code>arr</code> and an integer <code>difference</code>, return the length of the longest subsequence in <code>arr</code> which is an arithmetic sequence such that the difference between adjacent elements in the subsequence equals <code>difference</code>.</p>

<p>A <strong>subsequence</strong> is a sequence that can be derived from <code>arr</code> by deleting some or no elements without changing the order of the remaining elements.</p>

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

<pre>
<strong>Input:</strong> arr = [1,2,3,4], difference = 1
<strong>Output:</strong> 4
<strong>Explanation: </strong>The longest arithmetic subsequence is [1,2,3,4].</pre>

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

<pre>
<strong>Input:</strong> arr = [1,3,5,7], difference = 1
<strong>Output:</strong> 1
<strong>Explanation: </strong>The longest arithmetic subsequence is any single element.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,5,7,8,5,3,4,2,1], difference = -2
<strong>Output:</strong> 4
<strong>Explanation: </strong>The longest arithmetic subsequence is [7,5,3,1].
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= arr[i], difference &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int f(int idx,int p_idx,int d,vector<int>& ar,vector<vector<int>>& dp){
        if(idx==ar.size()) return 0;
        if(dp[idx][p_idx+1] != -1) return dp[idx][p_idx+1];
        int nt = f(idx+1,p_idx,d,ar,dp);
        int t = -1;
        if(p_idx == -1 || ar[idx]-ar[p_idx] == d){
            t = 1+f(idx+1,idx,d,ar,dp);
        }
        return dp[idx][p_idx+1] = max(t,nt);
    }
    int longestSubsequence(vector<int>& ar, int d) {
        int n = ar.size();
        unordered_map<int,int> dp;
        int ans = 1;
        for(int i=0;i<n;i++){
            int currEl = ar[i];
            int prevEl = ar[i]-d;
            dp[currEl] = dp[prevEl]+1;
            ans = max(ans,dp[currEl]);
        }
        return ans;
    }
};
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