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Count the Arithmetic sequences in the Array of size at least 3
  • Last Updated : 01 Jun, 2020

Given an array arr[] of size N, the task is to find the count of all arithmetic sequences in the array.

Examples:

Input: arr = [1, 2, 3, 4]
Output: 3
Explanation:
The arithmetic sequences in arr are [1, 2, 3], [2, 3, 4] and [1, 2, 3, 4] itself.

Input: arr = [1, 3, 5, 6, 7, 8]
Output: 4
Explanation:
The arithmetic sequences in arr are [1, 3, 5], [5, 6, 7], [5, 6, 7, 8] and [6, 7, 8].

Naive approach:



  • Run two loops and check for each sequence of length at least 3.
  • If the sequence is an arithmetic sequence, then increment the answer by 1.
  • Finally, return the count of all the arithmetic subarray of size at least 3.

Time Complexity: O(N2)

Efficient approach: We will use a dynamic programming approach to maintain a count of all arithmetic sequences till any position.

  • Initialize a variable, res with 0. It will store the count of sequences.
  • Initialize a variable, count with 0. It will store the size of the sequence minus 2.
  • Increase the value of count if the current element forms an arithmetic sequence else make it zero.
  • If the current element L[i] is making an arithmetic sequence with L[i-1] and L[i-2], then the number of arithmetic sequences till the ith iteration is given by:

    res = res + count

  • Finally, return the res variable.

Below is the implementation of the above approach:

C++




// C++ program to find all arithmetic
// sequences of size atleast 3
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find all arithmetic
// sequences of size atleast 3
int numberOfArithmeticSequences(int L[], int N)
{
  
    // If array size is less than 3
    if (N <= 2)
        return 0;
  
    // Finding arithmetic subarray length
    int count = 0;
  
    // To store all arithmetic subarray
    // of length at least 3
    int res = 0;
  
    for (int i = 2; i < N; ++i) {
  
        // Check if current element makes
        // atithmetic sequence with
        // previous two elements
        if (L[i] - L[i - 1] == L[i - 1] - L[i - 2]) {
            ++count;
        }
  
        // Begin with a new element for
        // new arithmetic sequences
        else {
            count = 0;
        }
  
        // Accumulate result in till i.
        res += count;
    }
  
    // Return final count
    return res;
}
  
// Driver code
int main()
{
  
    int L[] = { 1, 3, 5, 6, 7, 8 };
    int N = sizeof(L) / sizeof(L[0]);
  
    // Function to find arithematic sequences
    cout << numberOfArithmeticSequences(L, N);
  
    return 0;
}

Java




// Java program to find all arithmetic
// sequences of size atleast 3
  
class GFG{
   
// Function to find all arithmetic
// sequences of size atleast 3
static int numberOfArithmeticSequences(int L[], int N)
{
   
    // If array size is less than 3
    if (N <= 2)
        return 0;
   
    // Finding arithmetic subarray length
    int count = 0;
   
    // To store all arithmetic subarray
    // of length at least 3
    int res = 0;
   
    for (int i = 2; i < N; ++i) {
  
        // Check if current element makes
        // atithmetic sequence with
        // previous two elements
        if (L[i] - L[i - 1] == L[i - 1] - L[i - 2]) {
            ++count;
        }
   
        // Begin with a new element for
        // new arithmetic sequences
        else {
            count = 0;
        }
   
        // Accumulate result in till i.
        res += count;
    }
   
    // Return final count
    return res;
}
   
// Driver code
public static void main(String[] args)
{
   
    int L[] = { 1, 3, 5, 6, 7, 8 };
    int N = L.length;
   
    // Function to find arithmetic sequences
    System.out.print(numberOfArithmeticSequences(L, N));
   
}
}
  
// This code contributed by sapnasingh4991

Python3




# Python3 program to find all arithmetic 
# sequences of size atleast 3 
  
# Function to find all arithmetic 
# sequences of size atleast 3 
def numberOfArithmeticSequences(L, N) :
  
    # If array size is less than 3 
    if (N <= 2) :
        return 0
  
    # Finding arithmetic subarray length 
    count = 0
  
    # To store all arithmetic subarray 
    # of length at least 3 
    res = 0
  
    for i in range(2,N): 
  
        # Check if current element makes 
        # atithmetic sequence with 
        # previous two elements 
        if ( (L[i] - L[i - 1]) == (L[i - 1] - L[i - 2])) :
            count += 1
  
        # Begin with a new element for 
        # new arithmetic sequences 
        else :
            count = 0
  
        # Accumulate result in till i. 
        res += count
  
  
    # Return final count 
    return res
  
# Driver code 
  
L = [ 1, 3, 5, 6, 7, 8 ]
N = len(L)
  
# Function to find arithematic sequences 
print(numberOfArithmeticSequences(L, N))
  
# This code is contributed by Sanjit_Prasad

C#




// C# program to find all arithmetic
// sequences of size atleast 3
using System;
  
class GFG{
  
// Function to find all arithmetic
// sequences of size atleast 3
static int numberOfArithmeticSequences(int []L, 
                                       int N)
{
  
    // If array size is less than 3
    if (N <= 2)
        return 0;
  
    // Finding arithmetic subarray length
    int count = 0;
  
    // To store all arithmetic subarray
    // of length at least 3
    int res = 0;
  
    for(int i = 2; i < N; ++i)
    {
          
       // Check if current element makes
       // atithmetic sequence with
       // previous two elements
       if (L[i] - L[i - 1] ==
           L[i - 1] - L[i - 2])
       {
           ++count;
       }
         
       // Begin with a new element for
       // new arithmetic sequences
       else
       {
           count = 0;
       }
         
       // Accumulate result in till i.
       res += count;
    }
      
    // Return readonly count
    return res;
}
  
// Driver code
public static void Main(String[] args)
{
    int []L = { 1, 3, 5, 6, 7, 8 };
    int N = L.Length;
  
    // Function to find arithmetic sequences
    Console.Write(numberOfArithmeticSequences(L, N));
}
}
  
// This code is contributed by amal kumar choubey
Output:
4

Time Complexity: O(N)

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