Create 488.cpp

488.cpp

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+// C++ program to find Length of the Longest AP (llap) in a given sorted set.
+// The code strictly implements the algorithm provided in the reference.
+#include <iostream>
+using namespace std;
+
+// Returns length of the longest AP subset in a given set
+int lenghtOfLongestAP(int set[], int n)
+{
+	if (n <= 2) return n;
+
+	// Create a table and initialize all values as 2. The value of
+	// L[i][j] stores LLAP with set[i] and set[j] as first two
+	// elements of AP. Only valid entries are the entries where j>i
+	int L[n][n];
+	int llap = 2; // Initialize the result
+
+	// Fill entries in last column as 2. There will always be
+	// two elements in AP with last number of set as second
+	// element in AP
+	for (int i = 0; i < n; i++)
+		L[i][n-1] = 2;
+
+	// Consider every element as second element of AP
+	for (int j=n-2; j>=1; j--)
+	{
+		// Search for i and k for j
+		int i = j-1, k = j+1;
+		while (i >= 0 && k <= n-1)
+		{
+		if (set[i] + set[k] < 2*set[j])
+			k++;
+
+		// Before changing i, set L[i][j] as 2
+		else if (set[i] + set[k] > 2*set[j])
+		{ L[i][j] = 2, i--; }
+
+		else
+		{
+			// Found i and k for j, LLAP with i and j as first two
+			// elements is equal to LLAP with j and k as first two
+			// elements plus 1. L[j][k] must have been filled
+			// before as we run the loop from right side
+			L[i][j] = L[j][k] + 1;
+
+			// Update overall LLAP, if needed
+			llap = max(llap, L[i][j]);
+
+			// Change i and k to fill more L[i][j] values for
+			// current j
+			i--; k++;
+		}
+		}
+
+		// If the loop was stopped due to k becoming more than
+		// n-1, set the remaining entties in column j as 2
+		while (i >= 0)
+		{
+			L[i][j] = 2;
+			i--;
+		}
+	}
+	return llap;
+}
+
+/* Driver program to test above function*/
+int main()
+{
+	int set1[] = {1, 7, 10, 13, 14, 19};
+	int n1 = sizeof(set1)/sizeof(set1[0]);
+	cout << lenghtOfLongestAP(set1, n1) << endl;
+
+	int set2[] = {1, 7, 10, 15, 27, 29};
+	int n2 = sizeof(set2)/sizeof(set2[0]);
+	cout << lenghtOfLongestAP(set2, n2) << endl;
+
+	int set3[] = {2, 4, 6, 8, 10};
+	int n3 = sizeof(set3)/sizeof(set3[0]);
+	cout << lenghtOfLongestAP(set3, n3) << endl;
+
+	return 0;
+}