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# Longest subarray such that the difference of max and min is at-most one

• Difficulty Level : Hard
• Last Updated : 26 Aug, 2021

Given an array of n numbers where the difference between each number and the previous one doesn’t exceed one. Find the longest contiguous subarray such that the difference between the maximum and minimum number in the range doesn’t exceed one.

Examples:

```Input : {3, 3, 4, 4, 5, 6}
Output : 4
The longest subarray here is {3, 3, 4, 4}

Input : {7, 7, 7}
Output : 3
The longest subarray here is {7, 7, 7}

Input : {9, 8, 8, 9, 9, 10}
Output : 5
The longest subarray here is {9, 8, 8, 9, 9}```

If the difference between the maximum and the minimum numbers in the sequence doesn’t exceed one then the sequence consisted of only one or two consecutive numbers. The idea is to traverse the array and keep track of current element and previous element in current subarray. If we find an element which is not same as current or previous, we update current and previous. We also update result if required.

## C++

 `// C++ code to find longest``// subarray with difference``// between max and min as at-most 1.``#include``using` `namespace` `std;`` ` `int` `longestSubarray(``int` `input[], ``                    ``int` `length)``{``    ``int` `prev = -1;``    ``int` `current, next;``    ``int` `prevCount = 0, currentCount = 1;`` ` `    ``// longest constant range length``    ``int` `longest = 1;`` ` `    ``// first number``    ``current = input;`` ` `    ``for` `(``int` `i = 1; i < length; i++)``    ``{``        ``next = input[i];`` ` `        ``// If we see same number``        ``if` `(next == current) ``        ``{``            ``currentCount++;``        ``}`` ` `        ``// If we see different number, ``        ``// but same as previous.``        ``else` `if` `(next == prev)``        ``{``            ``prevCount += currentCount;``            ``prev = current;``            ``current = next;``            ``currentCount = 1;``        ``}`` ` `        ``// If number is neither same ``        ``// as previous nor as current. ``        ``else` `        ``{``            ``longest = max(longest, ``                          ``currentCount + prevCount);``            ``prev = current;``            ``prevCount = currentCount;``            ``current = next;``            ``currentCount = 1;``        ``}``    ``}`` ` `    ``return` `max(longest, ``               ``currentCount + prevCount);``}`` ` `// Driver Code``int` `main()``{``    ``int` `input[] = { 5, 5, 6, 7, 6 }; ``    ``int` `n = ``sizeof``(input) / ``sizeof``(``int``);``    ``cout << longestSubarray(input, n);``    ``return` `0;``}`` ` `// This code is contributed``// by Arnab Kundu`

## Java

 `// Java code to find longest subarray with difference``// between max and min as at-most 1.``public` `class` `Demo {`` ` `    ``public` `static` `int` `longestSubarray(``int``[] input)``    ``{``        ``int` `prev = -``1``;``        ``int` `current, next;``        ``int` `prevCount = ``0``, currentCount = ``1``;`` ` `        ``// longest constant range length``        ``int` `longest = ``1``;`` ` `        ``// first number``        ``current = input[``0``];`` ` `        ``for` `(``int` `i = ``1``; i < input.length; i++) {``            ``next = input[i];`` ` `            ``// If we see same number``            ``if` `(next == current) {``                ``currentCount++;``            ``}`` ` `            ``// If we see different number, but``            ``// same as previous.``            ``else` `if` `(next == prev) {``                ``prevCount += currentCount;``                ``prev = current;``                ``current = next;``                ``currentCount = ``1``;``            ``}`` ` `            ``// If number is neither same as previous``            ``// nor as current. ``            ``else` `{``                ``longest = Math.max(longest, currentCount + prevCount);``                ``prev = current;``                ``prevCount = currentCount;``                ``current = next;``                ``currentCount = ``1``;``            ``}``        ``}`` ` `        ``return` `Math.max(longest, currentCount + prevCount);``    ``}`` ` `    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] input = { ``5``, ``5``, ``6``, ``7``, ``6` `};``        ``System.out.println(longestSubarray(input));``    ``}``}`

## Python 3

 `# Python 3 code to find longest``# subarray with difference``# between max and min as at-most 1.``def` `longestSubarray(``input``, length):`` ` `    ``prev ``=` `-``1``    ``prevCount ``=` `0``    ``currentCount ``=` `1`` ` `    ``# longest constant range length``    ``longest ``=` `1`` ` `    ``# first number``    ``current ``=` `input``[``0``]`` ` `    ``for` `i ``in` `range``(``1``, length):`` ` `        ``next` `=` `input``[i]`` ` `        ``# If we see same number``        ``if` `next` `=``=` `current :``            ``currentCount ``+``=` `1``     ` `        ``# If we see different number, ``        ``# but same as previous.``        ``elif` `next` `=``=` `prev :``            ``prevCount ``+``=` `currentCount``            ``prev ``=` `current``            ``current ``=` `next``            ``currentCount ``=` `1``         ` `        ``# If number is neither same ``        ``# as previous nor as current. ``        ``else``:``            ``longest ``=` `max``(longest, ``                          ``currentCount ``+` `                          ``prevCount)``            ``prev ``=` `current``            ``prevCount ``=` `currentCount``            ``current ``=` `next``            ``currentCount ``=` `1``         ` `    ``return` `max``(longest, ``            ``currentCount ``+` `prevCount)`` ` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``input` `=` `[ ``5``, ``5``, ``6``, ``7``, ``6` `]``    ``n ``=` `len``(``input``)``    ``print``(longestSubarray(``input``, n))``     ` `# This code is contributed``# by ChitraNayal`

## C#

 `// C# code to find longest ``// subarray with difference ``// between max and min as ``// at-most 1.``using` `System;`` ` `class` `GFG``{``public` `static` `int` `longestSubarray(``int``[] input)``{``    ``int` `prev = -1;``    ``int` `current, next;``    ``int` `prevCount = 0, ``        ``currentCount = 1;`` ` `    ``// longest constant ``    ``// range length``    ``int` `longest = 1;`` ` `    ``// first number``    ``current = input;`` ` `    ``for` `(``int` `i = 1; ``             ``i < input.Length; i++)``    ``{``        ``next = input[i];`` ` `        ``// If we see same number``        ``if` `(next == current) ``        ``{``            ``currentCount++;``        ``}`` ` `        ``// If we see different number, ``        ``// but same as previous.``        ``else` `if` `(next == prev) ``        ``{``            ``prevCount += currentCount;``            ``prev = current;``            ``current = next;``            ``currentCount = 1;``        ``}`` ` `        ``// If number is neither ``        ``// same as previous``        ``// nor as current. ``        ``else` `        ``{``            ``longest = Math.Max(longest, ``                               ``currentCount + ``                               ``prevCount);``            ``prev = current;``            ``prevCount = currentCount;``            ``current = next;``            ``currentCount = 1;``        ``}``    ``}`` ` `    ``return` `Math.Max(longest, ``                    ``currentCount + prevCount);``}`` ` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int``[] input = {5, 5, 6, 7, 6};``    ``Console.WriteLine(longestSubarray(input));``}``}`` ` `// This code is contributed``// by Kirti_Mangal`

## PHP

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## Javascript

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Output:

`3`

Time Complexity: O(n) where n is the length of the input array.

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