# Length of Longest subarray such that difference between adjacent elements is K

• Last Updated : 21 Jan, 2022

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 =1
Output: 2
Explanation: 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 = 2
Output: 3
Explanation: 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 = 1
Output: 1
Explanation: 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 wasn’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 ``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);``    ``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

 ``

Output
`1`

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

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