# Longest subarray forming an Arithmetic Progression (AP) with given common difference

Given an array arr[] of N integers and an integer K, the task is to find the length of the longest subarray that forms an Arithmetic Progression having common difference K.

Examples:

Input: arr[] = {3, 4, 5}, K = 1
Output: 3
Explanation: The longest subarray forming an AP with common difference 1 is {3, 4, 5}.

Input: arr[] = {10, 7, 4, 6, 8, 10, 11}, K = 2
Output: 4
Explanation: The longest possible subarray forming an AP with common difference as 2 is {4, 6, 8, 10} .

Brute Force Approach: This approach finds the length of the longest subarray forming an arithmetic progression with the given common difference by looping through each possible subarray in the array and checking if it forms an arithmetic progression with the given common difference. It keeps track of the length of the current subarray and updates the answer if the length of the current subarray is greater than the current maximum length.

• Initialize a variable maxLen to 1.
• Loop through each element arr[i] in the array from index 0 to n-1.
• For each arr[i], initialize a variable count to 1 and a variable prev to arr[i].
• Loop through each element arr[j] in the array from index i+1 to n-1.
• Check if arr[j] prev is equal to the given common difference. If it is, increment count and set prev to arr[j].
• If arr[j] – prev is not equal to the given common difference, break out of the inner loop.
• After the inner loop, update maxLen to be maximum of maxLen and count.
• After the outer loop, return maxLen as the length of the longest subarray forming an arithmetic progression with the given common difference.

Below is the implementation of the above approach:

## C++

 #include using namespace std; // Function to find longest subarray// forming an Arithmetic Progression// with the given common differenceint maxlenAP(int arr[], int& n, int& d){    // Stores final answer    int maxLen = 1;      // Loop to traverse array    for (int i = 0; i < n; i++) {        // Stores the length of the        // current window        int count = 1;        int prev = arr[i];          // Loop to find the longest subarray        // starting at i        for (int j = i + 1; j < n; j++) {            if (arr[j] - prev == d) {                count++;                prev = arr[j];            }        }          // Update answer        maxLen = max(maxLen, count);    }      // Return Answer    return maxLen;} // Driver Codeint main(){    int arr[] = { 10, 7, 4, 6, 8, 10, 11 };    int N = sizeof(arr) / sizeof(arr[0]);    int K = 2;      cout << maxlenAP(arr, N, K);      return 0;}

## Java

 // Java program for// the above approachimport java.util.Arrays; public class GFG {     // Function to find longest subarray    // forming an Arithmetic Progression    // with the given common difference    static int maxlenAP(int[] arr, int n, int d)    {        // Stores final answer        int maxLen = 1;         // Loop to traverse array        for (int i = 0; i < n; i++) {            // Stores the length of the            // current window            int count = 1;            int prev = arr[i];             // Loop to find the longest subarray            // starting at i            for (int j = i + 1; j < n; j++) {                if (arr[j] - prev == d) {                    count++;                    prev = arr[j];                }            }             // Update answer            maxLen = Math.max(maxLen, count);        }         // Return Answer        return maxLen;    }     // Driver Code    public static void main(String[] args)    {        // Given array        int[] arr = { 10, 7, 4, 6, 8, 10, 11 };        int N = arr.length;        // Given subarray size K        int K = 2;        // Function Call        System.out.println(maxlenAP(arr, N, K));    }}

## Python

 # Function to find longest subarray# forming an Arithmetic Progression# with the given common difference def maxlenAP(arr, n, d):       # Stores the final answer    maxLen = 1     # Loop to traverse the array    for i in range(n):        # Stores the length of the current window        count = 1        prev = arr[i]         # Loop to find the longest subarray starting at i        for j in range(i + 1, n):            if arr[j] - prev == d:                count += 1                prev = arr[j]         # Update the answer        maxLen = max(maxLen, count)     # Return the answer    return maxLen # Driver Code  def main():    arr = [10, 7, 4, 6, 8, 10, 11]    N = len(arr)    K = 2     # Function call    print(maxlenAP(arr, N, K))  if __name__ == "__main__":    main()

## C#

 // C# program for// the above approach using System; public class GFG{    // Function to find longest subarray    // forming an Arithmetic Progression    // with the given common difference    public static int MaxLenAP(int[] arr, int n, int d)    {        // Stores the final answer        int maxLen = 1;         // Loop to traverse the array        for (int i = 0; i < n; i++)        {            // Stores the length of the current window            int count = 1;            int prev = arr[i];             // Loop to find the longest subarray starting at i            for (int j = i + 1; j < n; j++)            {                if (arr[j] - prev == d)                {                    count++;                    prev = arr[j];                }            }             // Update the answer            maxLen = Math.Max(maxLen, count);        }         // Return the answer        return maxLen;    }//Driver code    public static void Main()    {        int[] arr = { 10, 7, 4, 6, 8, 10, 11 };        int N = arr.Length;        int K = 2;         // Function call        Console.WriteLine(MaxLenAP(arr, N, K));    }}

## Javascript

 // Function to find longest subarray// forming an Arithmetic Progression// with the given common differencefunction maxlenAP(arr, n, d) {    // Stores final answer    let maxLen = 1;     // Loop to traverse array    for (let i = 0; i < n; i++) {        // Stores the length of the        // current window        let count = 1;        let prev = arr[i];         // Loop to find the longest subarray        // starting at i        for (let j = i + 1; j < n; j++) {            if (arr[j] - prev == d) {                count++;                prev = arr[j];            }        }         // Update answer        maxLen = Math.max(maxLen, count);    }     // Return Answer    return maxLen;} // Driver Codelet arr = [10, 7, 4, 6, 8, 10, 11];let N = arr.length;let K = 2; console.log(maxlenAP(arr, N, K));

Output:

4

Time Complexity: O(N2)
Auxiliary Space: O(1)

Sliding Window Approach: The given problem is an implementation-based problem that can be solved using the sliding window technique. Follow the steps mentioned below to solve the problem:

• Traverse the given array and maintain a variable that stores the number of variables in the current window.
• If the difference between the current element and the previous element in the array is K, increment the size of the current window, otherwise, reset the size of the window as 1.
• Print the maximum difference as answer.

Below is the implementation of the above approach:

## C++14

 // C++ program of the above approach#include using namespace std; // Function to find longest subarray// forming an Arithmetic Progression// with the given common differenceint maxlenAP(int arr[], int& n, int& d){    // Stores the length of    // the current window    int count = 1;     // Stores final answer    int maxLen = INT_MIN;     // Loop to traverse array    for (int i = 1; i < n; i++) {        if (arr[i] - arr[i - 1] == d)             // Increment window size            count++;        else             // Reset window size            count = 1;         // Update answer        maxLen = max(maxLen, count);    }     // Return Answer    return maxLen;} // Driver Codeint main(){    int arr[] = { 10, 7, 4, 6, 8, 10, 11 };    int N = sizeof(arr) / sizeof(arr[0]);    int K = 2;     cout << maxlenAP(arr, N, K);     return 0;}

## Java

 // Java program for the above approachimport java.util.*;public class GFG {     // Function to find longest subarray// forming an Arithmetic Progression// with the given common differencestatic int maxlenAP(int []arr, int n, int d){       // Stores the length of    // the current window    int count = 1;     // Stores final answer    int maxLen = Integer.MIN_VALUE;     // Loop to traverse array    for (int i = 1; i < n; i++) {        if (arr[i] - arr[i - 1] == d)             // Increment window size            count++;        else             // Reset window size            count = 1;         // Update answer        maxLen = Math.max(maxLen, count);    }     // Return Answer    return maxLen;} // Driver Codepublic static void main(String args[]){    int []arr = { 10, 7, 4, 6, 8, 10, 11 };    int N = arr.length;    int K = 2;     System.out.println(maxlenAP(arr, N, K)); }} // This code is contributed by Samim Hossain Mondal.

## Python3

 # Python code for the above approach # Function to find longest subarray# forming an Arithmetic Progression# with the given common differencedef maxlenAP(arr, n, d):     # Stores the length of    # the current window    count = 1     # Stores final answer    maxLen = 10 ** -9     # Loop to traverse array    for i in range(1, n):        if (arr[i] - arr[i - 1] == d):            # Increment window size            count += 1        else:            # Reset window size            count = 1         # Update answer        maxLen = max(maxLen, count)     # Return Answer    return maxLen  # Driver Codearr = [10, 7, 4, 6, 8, 10, 11]N = len(arr)K = 2print(maxlenAP(arr, N, K)) # This code is contributed by gfgking

## C#

 // C# program for the above approachusing System;class GFG {     // Function to find longest subarray// forming an Arithmetic Progression// with the given common differencestatic int maxlenAP(int []arr, int n, int d){       // Stores the length of    // the current window    int count = 1;     // Stores final answer    int maxLen = Int32.MinValue;     // Loop to traverse array    for (int i = 1; i < n; i++) {        if (arr[i] - arr[i - 1] == d)             // Increment window size            count++;        else             // Reset window size            count = 1;         // Update answer        maxLen = Math.Max(maxLen, count);    }     // Return Answer    return maxLen;} // Driver Codepublic static void Main(){    int []arr = { 10, 7, 4, 6, 8, 10, 11 };    int N = arr.Length;    int K = 2;     Console.Write(maxlenAP(arr, N, K)); }} // This code is contributed by Samim Hossain Mondal.

## Javascript



Output
4

Time Complexity: O(N)
Auxiliary Space: O(1)

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