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 implementation of above idea :
C++
// 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 sum 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; } |
Java
// Java program to find number of AP // subsequences in the given array import java.util.Arrays; class GFG { static final int MAX = 1000001; static int numofAP(int a[], int n) { // initializing the minimum value and // maximum value of the array. int minarr = +2147483647; int maxarr = -2147483648; // Finding the minimum and maximum // value of the array. for (int i = 0; i < n; i++) { minarr = Math.min(minarr, a[i]); maxarr = Math.max(maxarr, a[i]); } // dp[i] is going to store count of // APs ending with arr[i]. // sum[j] is going to store sum of // all dp[]'s with j as an AP element. int dp[] = new int[n]; int sum[] = new int[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++) { Arrays.fill(sum, 0); // 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 public static void main(String[] args) { int arr[] = { 1, 2, 3 }; int n = arr.length; System.out.println(numofAP(arr, n)); } } // This code is contributed by Anant Agarwal. |
Python3
# Python program to find number of AP # subsequences in the given array MAX = 1000001 def numofAP(a, n): # initializing the minimum value and # maximum value of the array. minarr = +2147483647 maxarr = -2147483648 # Finding the minimum and # maximum value of the array. for i in range(n): 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 sum of all dp[]'s # with j as an AP element. dp = [0 for i in range(n + 1)] # Initialize answer with n + 1 as single # elements and empty array are also DP. ans = n + 1 # Traversing with all common difference. for d in range((minarr - maxarr), (maxarr - minarr) + 1): sum = [0 for i in range(MAX + 1)] # Traversing all the element of the array. for i in range(n): # 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 and a[i] - d <= 1000000): dp[i] += sum[a[i] - d] ans += dp[i] - 1 sum[a[i]] += dp[i] return ans # Driver code arr = [ 1, 2, 3 ] n = len(arr) print(numofAP(arr, n)) # This code is contributed by Anant Agarwal. |
C#
// C# program to find number of AP // subsequences in the given array using System; class GFG { static int MAX = 1000001; // Function to find number of AP // subsequences in the given array static int numofAP(int []a, int n) { // initializing the minimum value and // maximum value of the array. int minarr = +2147483647; int maxarr = -2147483648; int i; // Finding the minimum and maximum // value of the array. for (i = 0; i < n; i++) { minarr = Math.Min(minarr, a[i]); maxarr = Math.Max(maxarr, a[i]); } // dp[i] is going to store count of // APs ending with arr[i]. // sum[j] is going to store sum of // all dp[]'s with j as an AP element. int []dp = new int[n]; int []sum = new int[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++) { for(i = 0; i < MAX; i++) sum[i]= 0; // Traversing all the element // of the array. for ( 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 public static void Main() { int []arr = {1, 2, 3}; int n = arr.Length; Console.WriteLine(numofAP(arr, n)); } } // This code is contributed by vt_m. |
Javascript
<script> // Javascript program to find number of AP // subsequences in the given array let MAX = 1000001; function numofAP(a, n) { // initializing the minimum value and // maximum value of the array. let minarr = +2147483647; let maxarr = -2147483648; // Finding the minimum and maximum // value of the array. for (let i = 0; i < n; i++) { minarr = Math.min(minarr, a[i]); maxarr = Math.max(maxarr, a[i]); } // dp[i] is going to store count of // APs ending with arr[i]. // sum[j] is going to store sum of // all dp[]'s with j as an AP element. let dp = new Array(n); let sum = new Array(MAX); // Initialize answer with n + 1 as // single elements and empty array // are also DP. let ans = n + 1; // Traversing with all common // difference. for (let d = (minarr - maxarr); d <= (maxarr - minarr); d++) { sum.fill(0); // Traversing all the element // of the array. for (let 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; } let arr = [ 1, 2, 3 ]; let n = arr.length; document.write(numofAP(arr, n)); </script> |
Output :
8
Time complexity: O(n*d) given an array of n positive integers and d is the difference between the maximum and minimum value in the array.
Auxiliary Space: O(MAX)
This article is contributed by
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