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

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}.```

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++ 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;`
`}`

Output:

```Maximum length subsequence = 5
```

Time Complexity: O(n2)
Auxiliary Space: O(n)

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rakesh