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# Maximum length subarray with difference between adjacent elements as either 0 or 1

• Difficulty Level : Easy
• Last Updated : 23 Jul, 2022

Given an array of n integers. The task is to find the maximum length of the sub-array such that absolute difference between all the consecutive elements of the sub-array is either 0 or 1.
Examples:

Input: arr[] = {2, 5, 6, 3, 7, 6, 5, 8}
Output:
{5, 6} and {7, 6, 5} are the only valid sub-arrays.
Input: arr[] = {-2, -1, 5, -1, 4, 0, 3}
Output:

Approach: Starting from the first element of the array, find the first valid sub-array and store it’s 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 maximum 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 either 1 or 0``int` `getMaxLength(``int` `arr[],``int` `n)``{``    ``int` `l = n;``    ``int` `i = 0, maxlen = 0;``    ``while` `(i < l)``    ``{``        ``int` `j = i;``        ``while` `(i+1 < l &&``             ``(``abs``(arr[i] - arr[i + 1]) == 1 ||``             ``abs``(arr[i] - arr[i + 1]) == 0))``        ``{``            ``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++;``    ``}` `    ``// Any valid sub-array cannot be of length 1``    ``//maxlen = (maxlen == 1) ? 0 : maxlen;` `    ``// Return the maximum possible length``    ``return` `maxlen;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 2, 4 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << getMaxLength(arr, n);``}` `// This code is contributed by``// Surendra_Gangwar`

## Java

 `// Java implementation of the approach``public` `class` `GFG {` `    ``// Function to return the maximum length``    ``// of the sub-array such that the``    ``// absolute difference between every two``    ``// consecutive elements is either 1 or 0``    ``public` `static` `int` `getMaxLength(``int` `arr[])``    ``{` `        ``int` `l = arr.length;``        ``int` `i = ``0``, maxlen = ``0``;``        ``while` `(i < l) {``            ``int` `j = i;``            ``while` `(i + ``1` `< l``                   ``&& (Math.abs(arr[i] - arr[i + ``1``]) == ``1``                       ``|| Math.abs(arr[i] - arr[i + ``1``]) == ``0``)) {``                ``i++;``            ``}` `            ``// Length of the valid sub-array currently``            ``// under cosideration``            ``int` `currLen = i - j + ``1``;` `            ``// Update the maximum length``            ``if` `(maxlen < currLen)``                ``maxlen = currLen;` `            ``if` `(j == i)``                ``i++;``        ``}` `        ``// Any valid sub-array cannot be of length 1``        ``maxlen = (maxlen == ``1``) ? ``0` `: maxlen;` `        ``// Return the maximum possible length``        ``return` `maxlen;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``2``, ``4` `};``        ``System.out.print(getMaxLength(arr));``    ``}``}`

## Python3

 `# Python3 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 either 1 or 0``def` `getMaxLength(arr, n) :``    ` `    ``l ``=` `n;``    ``i ``=` `0``; maxlen ``=` `0``;``    ` `    ``while` `(i < l) :``        ``j ``=` `i;``        ``while` `(i ``+` `1` `< l ``and``              ``(``abs``(arr[i] ``-` `arr[i ``+` `1``]) ``=``=` `1` `or``               ``abs``(arr[i] ``-` `arr[i ``+` `1``]) ``=``=` `0``)) :``        ` `            ``i ``+``=` `1``;``        ` `        ``# Length of the valid sub-array``        ``# currently under cosideration``        ``currLen ``=` `i ``-` `j ``+` `1``;` `        ``# Update the maximum length``        ``if` `(maxlen < currLen) :``            ``maxlen ``=` `currLen;` `        ``if` `(j ``=``=` `i) :``            ``i ``+``=` `1``;``    ` `    ``# Any valid sub-array cannot be of length 1``    ``# maxlen = (maxlen == 1) ? 0 : maxlen;` `    ``# Return the maximum possible length``    ``return` `maxlen;``    ` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr ``=` `[ ``2``, ``4` `];``    ``n ``=` `len``(arr)``    ``print``(getMaxLength(arr, n));` `# This code is contributed by Ryuga`

## C#

 `// C# implementation of the 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 either 1 or 0``    ``public` `static` `int` `getMaxLength(``int` `[]arr)``    ``{` `        ``int` `l = arr.Length;``        ``int` `i = 0, maxlen = 0;``        ``while` `(i < l)``        ``{``            ``int` `j = i;``            ``while` `(i + 1 < l &&``                    ``(Math.Abs(arr[i] - arr[i + 1]) == 1 ||``                    ``Math.Abs(arr[i] - arr[i + 1]) == 0))``            ``{``                ``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++;``        ``}` `        ``// Any valid sub-array cannot be of length 1``        ``maxlen = (maxlen == 1) ? 0 : maxlen;` `        ``// Return the maximum possible length``        ``return` `maxlen;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String []args)``    ``{``        ``int` `[]arr = { 2, 4 };``        ``Console.Write(getMaxLength(arr));``    ``}``}` `// This code is contributed by Arnab Kundu`

## PHP

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## Javascript

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Output:

`1`

Time Complexity : O(n) ,where n is size of given array.

Space Complexity : O(1) ,as we are not using any extra space.

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