<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 longestSubsequence(vector<int>& arr, int difference) {
        int n = arr.size();
        if (n == 0) return 0;

        int minVal = *min_element(arr.begin(), arr.end());
        int maxVal = *max_element(arr.begin(), arr.end());
        
        vector<int> dp(maxVal - minVal + 1, 0);
        int globalBest = 1;

        for (int num : arr) {
            int index = num - minVal;
            int prevIndex = index - difference;
            
            if (prevIndex >= 0 && prevIndex < dp.size()) {
                dp[index] = dp[prevIndex] + 1;
            } else {
                dp[index] = 1;
            }
            
            globalBest = max(globalBest, dp[index]);
        }

        return globalBest;
    }
};
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