Given an array of **n** positive integers. The task is to count the number of Arithmetic Progression subsequence in the array. Note: Empty sequence or single element sequence is Arithmetic Progression. 1 <= arr[i] <= 1000000.

Examples:

Input : arr[] = { 1, 2, 3 } Output : 8 Arithmetic Progression subsequence from the given array are: {}, { 1 }, { 2 }, { 3 }, { 1, 2 }, { 2, 3 }, { 1, 3 }, { 1, 2, 3 }. Input : arr[] = { 10, 20, 30, 45 } Output : 12 Input : arr[] = { 1, 2, 3, 4, 5 } Output : 23

Since empty sequence and single element sequence is also arithmetic progression, so we initialize the answer with n(number of element in the array) + 1.

Now, we need to find the arithmetic progression subsequence of length greater than or equal to 2. Let minimum and maximum of the array be minarr and maxarr respectively. Observe, in all the arithmetic progression subsequences, the range of common difference will be from (minarr – maxarr) to (maxarr – minarr). Now, for each common difference, say d, calculate the subsequence of length greater than or equal to 2 using dynamic programming.

Let dp[i] be the number of subsequence that end with arr[i] and have common difference of d. So,

The number of subsequence of length greater than or equal to 2 with common difference d is sum of dp[i] – 1, 0 <= i = 2 with difference d. To speed up, store the sum of dp[j] with arr[j] + d = arr[i] and j < i.

Below is C++ implementation of this approach:

// C++ program to find number of AP // subsequences in the given array #include<bits/stdc++.h> #define MAX 1000001 using namespace std; int numofAP(int a[], int n) { // initializing the minimum value and // maximum value of the array. int minarr = INT_MAX, maxarr = INT_MIN; // Finding the minimum and maximum // value of the array. for (int i = 0; i < n; i++) { minarr = min(minarr, a[i]); maxarr = max(maxarr, a[i]); } // dp[i] is going to store count of APs ending // with arr[i]. // sum[j] is going to store sun of all dp[]'s // with j as an AP element. int dp[n], sum[MAX]; // Initialize answer with n + 1 as single elements // and empty array are also DP. int ans = n + 1; // Traversing with all common difference. for (int d=(minarr-maxarr); d<=(maxarr-minarr); d++) { memset(sum, 0, sizeof sum); // Traversing all the element of the array. for (int i = 0; i < n; i++) { // Initialize dp[i] = 1. dp[i] = 1; // Adding counts of APs with given differences // and a[i] is last element. // We consider all APs where an array element // is previous element of AP with a particular // difference if (a[i] - d >= 1 && a[i] - d <= 1000000) dp[i] += sum[a[i] - d]; ans += dp[i] - 1; sum[a[i]] += dp[i]; } } return ans; } // Driver code int main() { int arr[] = { 1, 2, 3 }; int n = sizeof(arr)/sizeof(arr[0]); cout << numofAP(arr, n) << endl; return 0; }

Output:

8

This article is contributed by **Anuj Chauhan**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.