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.

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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