Given an array **arr[]** of size **N**, and integer **K**. The task is to find the length of the longest subarray with the difference between adjacent elements as **K**.

**Examples:**

Input:arr[] = { 5, 5, 5, 10, 8, 6, 12, 13 }, K =1Output:2Explanation:Only one subarray which have difference between adjacents as 1 is {12, 13}.

Input:arr[] = {4, 6, 8, 9, 8, 12, 14, 17, 15}, K = 2Output:3Explanation:There are three such subarrays {4, 6, 8}, {12, 14} and {17, 15}.

{4, 6, 8} has the highest length.

Input:arr[] = {2, 2, 4, 6}, K = 1Output:1Explanation:No subarray of length more than satisfies this criteria.

**Approach:** Starting from the first element of the array, find the first valid sub-array and store its length then starting from the next element (the first element that wasnâ€™t included in the first sub-array), find another valid sub-array. Repeat the process until all the valid sub-arrays have been found then print the length of the longest sub-array.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return the maximum length` `// of the sub-array such that the` `// absolute difference between every two` `// consecutive elements is K` `int` `getMaxLength(` `int` `arr[], ` `int` `N, ` `int` `K)` `{` ` ` `int` `l = N;` ` ` `int` `i = 0, maxlen = 0;` ` ` `while` `(i < l) {` ` ` `int` `j = i;` ` ` `while` `(i + 1 < l` ` ` `&& (` `abs` `(arr[i] - ` ` ` `arr[i + 1]) == K)) {` ` ` `i++;` ` ` `}` ` ` `// Length of the valid sub-array ` ` ` `// currently under consideration` ` ` `int` `currLen = i - j + 1;` ` ` `// Update the maximum length` ` ` `if` `(maxlen < currLen)` ` ` `maxlen = currLen;` ` ` `if` `(j == i)` ` ` `i++;` ` ` `}` ` ` `// Return the maximum possible length` ` ` `return` `maxlen;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 2, 2, 4, 6 };` ` ` `int` `K = 1;` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `cout << getMaxLength(arr, N, K);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.util.*;` `public` `class` `GFG` `{` `// Function to return the maximum length` `// of the sub-array such that the` `// absolute difference between every two` `// consecutive elements is K` `static` `int` `getMaxLength(` `int` `arr[], ` `int` `N, ` `int` `K)` `{` ` ` `int` `l = N;` ` ` `int` `i = ` `0` `, maxlen = ` `0` `;` ` ` `while` `(i < l) {` ` ` `int` `j = i;` ` ` `while` `(i + ` `1` `< l` ` ` `&& (Math.abs(arr[i] - ` ` ` `arr[i + ` `1` `]) == K)) {` ` ` `i++;` ` ` `}` ` ` `// Length of the valid sub-array ` ` ` `// currently under consideration` ` ` `int` `currLen = i - j + ` `1` `;` ` ` `// Update the maximum length` ` ` `if` `(maxlen < currLen)` ` ` `maxlen = currLen;` ` ` `if` `(j == i)` ` ` `i++;` ` ` `}` ` ` `// Return the maximum possible length` ` ` `return` `maxlen;` `}` `// Driver code` `public` `static` `void` `main(String args[])` `{` ` ` `int` `arr[] = { ` `2` `, ` `2` `, ` `4` `, ` `6` `};` ` ` `int` `K = ` `1` `;` ` ` `int` `N = arr.length; ` ` ` `System.out.print(getMaxLength(arr, N, K));` `}` `}` `// This code is contributed by Samim Hossain Mondal.` |

## Python3

`# Python implementation of the approach` `# Function to return the maximum length` `# of the sub-array such that the` `# absolute difference between every two` `# consecutive elements is K` `def` `getMaxLength (arr, N, K):` ` ` `l ` `=` `N;` ` ` `i ` `=` `0` ` ` `maxlen ` `=` `0` `;` ` ` `while` `(i < l):` ` ` `j ` `=` `i;` ` ` `while` `(i ` `+` `1` `< l ` `and` `(` `abs` `(arr[i] ` `-` `arr[i ` `+` `1` `]) ` `=` `=` `K)):` ` ` `i ` `+` `=` `1` ` ` `# Length of the valid sub-array` ` ` `# currently under consideration` ` ` `currLen ` `=` `i ` `-` `j ` `+` `1` `;` ` ` `# Update the maximum length` ` ` `if` `(maxlen < currLen):` ` ` `maxlen ` `=` `currLen;` ` ` `if` `(j ` `=` `=` `i):` ` ` `i ` `+` `=` `1` ` ` `# Return the maximum possible length` ` ` `return` `maxlen;` `# Driver code` `arr ` `=` `[` `2` `, ` `2` `, ` `4` `, ` `6` `];` `K ` `=` `1` `;` `N ` `=` `len` `(arr)` `print` `(getMaxLength(arr, N, K));` `# This code is contributed by gfgking` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG` `{` ` ` `// Function to return the maximum length` ` ` `// of the sub-array such that the` ` ` `// absolute difference between every two` ` ` `// consecutive elements is K` ` ` `static` `int` `getMaxLength(` `int` `[]arr, ` `int` `N, ` `int` `K)` ` ` `{` ` ` `int` `l = N;` ` ` `int` `i = 0, maxlen = 0;` ` ` `while` `(i < l) {` ` ` `int` `j = i;` ` ` `while` `(i + 1 < l` ` ` `&& (Math.Abs(arr[i] - ` ` ` `arr[i + 1]) == K)) {` ` ` `i++;` ` ` `}` ` ` `// Length of the valid sub-array ` ` ` `// currently under consideration` ` ` `int` `currLen = i - j + 1;` ` ` `// Update the maximum length` ` ` `if` `(maxlen < currLen)` ` ` `maxlen = currLen;` ` ` `if` `(j == i)` ` ` `i++;` ` ` `}` ` ` `// Return the maximum possible length` ` ` `return` `maxlen;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `[]arr = { 2, 2, 4, 6 };` ` ` `int` `K = 1;` ` ` `int` `N = arr.Length;` ` ` `Console.Write(getMaxLength(arr, N, K));` ` ` `}` `}` `// This code is contributed by Samim Hossain Mondal.` |

## Javascript

`<script>` ` ` `// JavaScript implementation of the approach` ` ` `// Function to return the maximum length` ` ` `// of the sub-array such that the` ` ` `// absolute difference between every two` ` ` `// consecutive elements is K` ` ` `const getMaxLength = (arr, N, K) => {` ` ` `let l = N;` ` ` `let i = 0, maxlen = 0;` ` ` `while` `(i < l) {` ` ` `let j = i;` ` ` `while` `(i + 1 < l` ` ` `&& (Math.abs(arr[i] -` ` ` `arr[i + 1]) == K)) {` ` ` `i++;` ` ` `}` ` ` `// Length of the valid sub-array` ` ` `// currently under consideration` ` ` `let currLen = i - j + 1;` ` ` `// Update the maximum length` ` ` `if` `(maxlen < currLen)` ` ` `maxlen = currLen;` ` ` `if` `(j == i)` ` ` `i++;` ` ` `}` ` ` `// Return the maximum possible length` ` ` `return` `maxlen;` ` ` `}` ` ` `// Driver code` ` ` `let arr = [2, 2, 4, 6];` ` ` `let K = 1;` ` ` `let N = arr.length;` ` ` `document.write(getMaxLength(arr, N, K));` `// This code is contributed by rakeshsahni` `</script>` |

**Output**

1

* Time Complexity:* O(N

^{2})

*O(1)*

**Auxiliary Space:**