# Longest non-decreasing subsequence having difference between adjacent elements less than D

• Last Updated : 27 Oct, 2021

Given an array arr[] of N integers and an integer D, the task is to find the length of the longest non-decreasing subsequence such that the difference between every adjacent element is less than D.

Examples:

Input: arr[] = {1, 3, 2, 4, 5}, D = 2
Output: 3
Explanation:
Consider the subsequence as {3, 4, 5}, which is of maximum length = 3 satisfying the given criteria.

Input: arr[] = {1, 5, 3, 2, 7}, D = 2
Output: 2

Approach: The given problem is a variation of Longest Increasing Subsequence with criteria for the difference between adjacent array elements as less than D, this idea can be implemented using Dynamic Programming. Follow the steps below to solve the given problem:

• Initialize a dp array, where dp[i] will store the maximum length of non-decreasing subsequence after including the ith element such that the difference between every adjacent pair of elements is less than D.
• Initialize all values of the array dp[] as 1.
• Iterate a loop over the range [0, N] and in each iteration, i traverse the given array arr[] over the range [0, i – 1] using the variable j and if the value of arr[j] is at least arr[i] and the difference between them is less than D, then update the value of dp[i] to the maximum of dp[i] and (1 + dp[j]).
• After completing the above steps, print the maximum value of the array dp[] as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to return the length of the``// longest non-decreasing subsequence``// having the difference as D for every``// adjacent elements``int` `longestSubsequence(vector<``int``> arr,``                       ``int` `d)``{``    ``// Store the size of array``    ``int` `n = arr.size();` `    ``// Stores the maximum length of the``    ``// subsequence after including the``    ``// ith element``    ``vector<``int``> dp(n, 1);` `    ``for` `(``int` `i = 0; i < n; i++) {``        ``for` `(``int` `j = 0; j < i; j++) {` `            ``// If it satisfies the``            ``// given condition``            ``if` `(arr[i] - d < arr[j]``                ``and arr[i] >= arr[j]) {` `                ``// Update dp[i]``                ``dp[i] = max(dp[i], dp[j] + 1);``            ``}``        ``}``    ``}` `    ``// Maximum value in the dp``    ``// table is the answer``    ``return` `*max_element(``        ``dp.begin(), dp.end());``}` `// Driver Code``int` `main()``{``    ``vector<``int``> arr = { 1, 3, 2, 4, 5 };``    ``int` `D = 2;``    ``cout << longestSubsequence(arr, D);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;``public` `class` `GFG {``    ` `    ``// Function to return the length of the``    ``// longest non-decreasing subsequence``    ``// having the difference as D for every``    ``// adjacent elements``    ``static` `int` `longestSubsequence(``int`  `[]arr,``                           ``int` `d)``    ``{``      ` `        ``// Store the size of array``        ``int` `n = arr.length;``    ` `        ``// Stores the maximum length of the``        ``// subsequence after including the``        ``// ith element``        ``int` `[]dp = ``new` `int``[n];``        ` `        ``for``(``int` `i = ``0``; i < n ; i++)``            ``dp[i] = ``1``;``    ` `        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``for` `(``int` `j = ``0``; j < i; j++) {``    ` `                ``// If it satisfies the``                ``// given condition``                ``if` `(arr[i] - d < arr[j] && arr[i] >= arr[j]) {``    ` `                    ``// Update dp[i]``                    ``dp[i] = Math.max(dp[i], dp[j] + ``1``);``                ``}``            ``}``        ``}``    ` `        ``// Maximum value in the dp``        ``// table is the answer``        ``Arrays.sort(dp);``        ``return` `dp[n - ``1``];``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main (String[] args) {``        ``int` `arr[] = { ``1``, ``3``, ``2``, ``4``, ``5` `};``        ``int` `D = ``2``;``        ``System.out.println(longestSubsequence(arr, D));``    ``}``}` `// This code is contributed by AnkThon`

## Python3

 `# python program for the above approach` `# Function to return the length of the``# longest non-decreasing subsequence``# having the difference as D for every``# adjacent elements``def` `longestSubsequence(arr, d):` `    ``# Store the size of array``    ``n ``=` `len``(arr)` `    ``# Stores the maximum length of the``    ``# subsequence after including the``    ``# ith element``    ``dp ``=` `[``1` `for` `_ ``in` `range``(n)]` `    ``for` `i ``in` `range``(``0``, n):``        ``for` `j ``in` `range``(``0``, i):` `            ``# If it satisfies the``            ``# given condition``            ``if` `(arr[i] ``-` `d < arr[j] ``and` `arr[i] >``=` `arr[j]):` `                ``# Update dp[i]``                ``dp[i] ``=` `max``(dp[i], dp[j] ``+` `1``)` `    ``# Maximum value in the dp``    ``# table is the answer``    ``return` `max``(dp)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[``1``, ``3``, ``2``, ``4``, ``5``]``    ``D ``=` `2``    ``print``(longestSubsequence(arr, D))` `    ``# This code is contributed by rakeshsahni`

## C#

 `// C# program for the above approach``using` `System;` `public` `class` `GFG {``    ` `    ``// Function to return the length of the``    ``// longest non-decreasing subsequence``    ``// having the difference as D for every``    ``// adjacent elements``    ``static` `int` `longestSubsequence(``int`  `[]arr,``                           ``int` `d)``    ``{``      ` `        ``// Store the size of array``        ``int` `n = arr.Length;``    ` `        ``// Stores the maximum length of the``        ``// subsequence after including the``        ``// ith element``        ``int` `[]dp = ``new` `int``[n];``        ` `        ``for``(``int` `i = 0; i < n ; i++)``            ``dp[i] = 1;``    ` `        ``for` `(``int` `i = 0; i < n; i++) {``            ``for` `(``int` `j = 0; j < i; j++) {``    ` `                ``// If it satisfies the``                ``// given condition``                ``if` `(arr[i] - d < arr[j] && arr[i] >= arr[j]) {``    ` `                    ``// Update dp[i]``                    ``dp[i] = Math.Max(dp[i], dp[j] + 1);``                ``}``            ``}``        ``}``    ` `        ``// Maximum value in the dp``        ``// table is the answer``        ``Array.Sort(dp);``        ``return` `dp[n - 1];``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main (``string``[] args) {``        ``int` `[]arr = { 1, 3, 2, 4, 5 };``        ``int` `D = 2;``        ``Console.WriteLine(longestSubsequence(arr, D));``    ``}``}` `// This code is contributed by AnkThon`

## Javascript

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Output:

`3`

Time Complexity: O(N2)
Auxiliary Space: O(N)

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