# 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}.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.
```

Output:

```Maximum length subsequence = 5
```

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

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

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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