Given an array of **n** integers. The problem is to find maximum length of the subsequence with difference between adjacent elements as either 0 or 1.

Examples:

Input : arr[] = {2, 5, 6, 3, 7, 6, 5, 8} Output : 5 The subsequence is {5, 6, 7, 6, 5}. Input : arr[] = {-2, -1, 5, -1, 4, 0, 3} Output : 4 The subsequence is {-2, -1, -1, 0}.

**Source:** Expedia Interview Experience | Set 12

The solution to this problem closely resembles the Longest Increasing Subsequence problem. The only difference is that here we have to check whether the absolute difference between the adjacent elements of the subsequence is either 0 or 1.

## C++

`// C++ implementation to find maximum length ` `// subsequence with difference between adjacent ` `// elements as either 0 or 1 ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// function to find maximum length subsequence ` `// with difference between adjacent elements as ` `// either 0 or 1 ` `int` `maxLenSub(` `int` `arr[], ` `int` `n) ` `{ ` ` ` `int` `mls[n], max = 0; ` ` ` ` ` `// Initialize mls[] values for all indexes ` ` ` `for` `(` `int` `i=0; i<n; i++) ` ` ` `mls[i] = 1; ` ` ` ` ` `// Compute optimized maximum length subsequence ` ` ` `// values in bottom up manner ` ` ` `for` `(` `int` `i=1; i<n; i++) ` ` ` `for` `(` `int` `j=0; j<i; j++) ` ` ` `if` `(` `abs` `(arr[i] - arr[j]) <= 1 && ` ` ` `mls[i] < mls[j] + 1) ` ` ` `mls[i] = mls[j] + 1; ` ` ` ` ` `// Store maximum of all 'mls' values in 'max' ` ` ` `for` `(` `int` `i=0; i<n; i++) ` ` ` `if` `(max < mls[i]) ` ` ` `max = mls[i]; ` ` ` ` ` `// required maximum length subsequence ` ` ` `return` `max; ` `} ` ` ` `// Driver program to test above ` `int` `main() ` `{ ` ` ` `int` `arr[] = {2, 5, 6, 3, 7, 6, 5, 8}; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `cout << ` `"Maximum length subsequence = "` ` ` `<< maxLenSub(arr, n); ` ` ` `return` `0; ` `} ` |

## Java

`// JAVA Code for Maximum length subsequence ` `// with difference between adjacent elements ` `// as either 0 or 1 ` `import` `java.util.*; ` ` ` `class` `GFG { ` ` ` ` ` `// function to find maximum length subsequence ` ` ` `// with difference between adjacent elements as ` ` ` `// either 0 or 1 ` ` ` `public` `static` `int` `maxLenSub(` `int` `arr[], ` `int` `n) ` ` ` `{ ` ` ` `int` `mls[] = ` `new` `int` `[n], max = ` `0` `; ` ` ` ` ` `// Initialize mls[] values for all indexes ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `mls[i] = ` `1` `; ` ` ` ` ` `// Compute optimized maximum length ` ` ` `// subsequence values in bottom up manner ` ` ` `for` `(` `int` `i = ` `1` `; i < n; i++) ` ` ` `for` `(` `int` `j = ` `0` `; j < i; j++) ` ` ` `if` `(Math.abs(arr[i] - arr[j]) <= ` `1` ` ` `&& mls[i] < mls[j] + ` `1` `) ` ` ` `mls[i] = mls[j] + ` `1` `; ` ` ` ` ` `// Store maximum of all 'mls' values in 'max' ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `if` `(max < mls[i]) ` ` ` `max = mls[i]; ` ` ` ` ` `// required maximum length subsequence ` ` ` `return` `max; ` ` ` `} ` ` ` ` ` `/* Driver program to test above function */` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `arr[] = {` `2` `, ` `5` `, ` `6` `, ` `3` `, ` `7` `, ` `6` `, ` `5` `, ` `8` `}; ` ` ` `int` `n = arr.length; ` ` ` `System.out.println(` `"Maximum length subsequence = "` `+ ` ` ` `maxLenSub(arr, n)); ` ` ` ` ` `} ` `} ` ` ` `// This code is contributed by Arnav Kr. Mandal. ` |

## Python3

`# Python implementation to find maximum length ` `# subsequence with difference between adjacent ` `# elements as either 0 or 1 ` ` ` `# function to find maximum length subsequence ` `# with difference between adjacent elements as ` `# either 0 or 1 ` `def` `maxLenSub( arr, n): ` ` ` `mls` `=` `[] ` ` ` `max` `=` `0` ` ` ` ` `#Initialize mls[] values for all indexes ` ` ` `for` `i ` `in` `range` `(n): ` ` ` `mls.append(` `1` `) ` ` ` ` ` `#Compute optimized maximum length subsequence ` ` ` `# values in bottom up manner ` ` ` `for` `i ` `in` `range` `(n): ` ` ` `for` `j ` `in` `range` `(i): ` ` ` `if` `(` `abs` `(arr[i] ` `-` `arr[j]) <` `=` `1` `and` `mls[i] < mls[j] ` `+` `1` `): ` ` ` `mls[i] ` `=` `mls[j] ` `+` `1` ` ` ` ` `# Store maximum of all 'mls' values in 'max' ` ` ` `for` `i ` `in` `range` `(n): ` ` ` `if` `(` `max` `< mls[i]): ` ` ` `max` `=` `mls[i] ` ` ` ` ` `#required maximum length subsequence ` ` ` `return` `max` ` ` `#Driver program to test above ` `arr ` `=` `[` `2` `, ` `5` `, ` `6` `, ` `3` `, ` `7` `, ` `6` `, ` `5` `, ` `8` `] ` `n ` `=` `len` `(arr) ` `print` `(` `"Maximum length subsequence = "` `,maxLenSub(arr, n)) ` ` ` `#This code is contributed by "Abhishek Sharma 44" ` |

## C#

`// C# Code for Maximum length subsequence ` `// with difference between adjacent elements ` `// as either 0 or 1 ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `// function to find maximum length subsequence ` ` ` `// with difference between adjacent elements as ` ` ` `// either 0 or 1 ` ` ` `public` `static` `int` `maxLenSub(` `int` `[] arr, ` `int` `n) ` ` ` `{ ` ` ` `int` `[] mls = ` `new` `int` `[n]; ` ` ` `int` `max = 0; ` ` ` ` ` `// Initialize mls[] values for all indexes ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `mls[i] = 1; ` ` ` ` ` `// Compute optimized maximum length ` ` ` `// subsequence values in bottom up manner ` ` ` `for` `(` `int` `i = 1; i < n; i++) ` ` ` `for` `(` `int` `j = 0; j < i; j++) ` ` ` `if` `(Math.Abs(arr[i] - arr[j]) <= 1 ` ` ` `&& mls[i] < mls[j] + 1) ` ` ` `mls[i] = mls[j] + 1; ` ` ` ` ` `// Store maximum of all 'mls' values in 'max' ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `if` `(max < mls[i]) ` ` ` `max = mls[i]; ` ` ` ` ` `// required maximum length subsequence ` ` ` `return` `max; ` ` ` `} ` ` ` ` ` `/* Driver program to test above function */` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `[] arr = { 2, 5, 6, 3, 7, 6, 5, 8 }; ` ` ` `int` `n = arr.Length; ` ` ` `Console.Write(` `"Maximum length subsequence = "` `+ ` ` ` `maxLenSub(arr, n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Sam007 ` |

## PHP

`<?php ` `// PHP implementation to find maximum length ` `// subsequence with difference between adjacent ` `// elements as either 0 or 1 ` ` ` `// function to find maximum length subsequence ` `// with difference between adjacent elements as ` `// either 0 or 1 ` `function` `maxLenSub(` `$arr` `, ` `$n` `) ` `{ ` ` ` `$mls` `= ` `array` `(); ` `$max` `= 0; ` ` ` ` ` `// Initialize mls[] values ` ` ` `// for all indexes ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++) ` ` ` `$mls` `[` `$i` `] = 1; ` ` ` ` ` `// Compute optimized maximum ` ` ` `// length subsequence ` ` ` `// values in bottom up manner ` ` ` `for` `(` `$i` `= 1; ` `$i` `< ` `$n` `; ` `$i` `++) ` ` ` `for` `( ` `$j` `= 0; ` `$j` `< ` `$i` `; ` `$j` `++) ` ` ` `if` `(` `abs` `(` `$arr` `[` `$i` `] - ` `$arr` `[` `$j` `]) <= 1 ` `and` ` ` `$mls` `[` `$i` `] < ` `$mls` `[` `$j` `] + 1) ` ` ` `$mls` `[` `$i` `] = ` `$mls` `[` `$j` `] + 1; ` ` ` ` ` `// Store maximum of all ` ` ` `// 'mls' values in 'max' ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++) ` ` ` `if` `(` `$max` `< ` `$mls` `[` `$i` `]) ` ` ` `$max` `= ` `$mls` `[` `$i` `]; ` ` ` ` ` `// required maximum ` ` ` `// length subsequence ` ` ` `return` `$max` `; ` `} ` ` ` ` ` `// Driver Code ` ` ` `$arr` `= ` `array` `(2, 5, 6, 3, 7, 6, 5, 8); ` ` ` `$n` `= ` `count` `(` `$arr` `); ` ` ` `echo` `"Maximum length subsequence = "` ` ` `, maxLenSub(` `$arr` `, ` `$n` `); ` ` ` `// This code is contributed by anuj_67. ` `?> ` |

Output:

Maximum length subsequence = 5

Time Complexity: O(n^{2})

Auxiliary Space: O(n)

Maximum length subsequence with difference between adjacent elements as either 0 or 1 | Set 2

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