Create 04 - Space Optimized | DP | Approach.cpp

Author: JawadSher

24 - Dynamic Programming Problems/26 - Longest Arithmetic Subsequence/04 - Space Optimized | DP | Approach.cpp

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+class Solution {
+  public:
+    // Function to find the length of the longest arithmetic progression (AP) in the array
+    int lengthOfLongestAP(vector<int>& arr) {
+        int n = arr.size();
+
+        // If the array has 2 or fewer elements, the entire array is an AP
+        if(n <= 2) return n;
+
+        int ans = 2; // Minimum length of an AP is 2
+
+        // `dp` is a 2D map where:
+        // - The first key represents the index `i` of the array
+        // - The second key represents the difference `diff`
+        // - The value is the length of the longest AP ending at `i` with common difference `diff`
+        unordered_map<int, unordered_map<int, int>> dp;
+
+        // Iterate through all pairs of elements to fill the `dp` map
+        for(int i = 1; i < n; i++) { // Start from the second element
+            for(int j = 0; j < i; j++) { // Compare with all previous elements
+                // Calculate the common difference between `arr[i]` and `arr[j]`
+                int diff = arr[i] - arr[j];
+
+                // Retrieve the length of the AP ending at `j` with the same `diff`
+                // If no such AP exists, initialize it with length 1 (current pair)
+                int count = dp[j][diff] ? dp[j][diff] : 1;
+
+                // Update the length of the AP ending at `i` with the same `diff`
+                dp[i][diff] = count + 1;
+
+                // Update the maximum length of AP found so far
+                ans = max(ans, dp[i][diff]);
+            }
+        }
+
+        // Return the maximum length of AP found
+        return ans;
+    }
+};