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

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 = 2Output:3Explanation:

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 = 2Output: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**i**element such that the difference between every adjacent pair of elements is less than^{th}**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 <bits/stdc++.h>` `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

`<script>` `// Javascript 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` `function` `longestSubsequence(arr, d)` `{` ` ` `// Store the size of array` ` ` `let n = arr.length;` ` ` `// Stores the maximum length of the` ` ` `// subsequence after including the` ` ` `// ith element` ` ` `let dp = ` `new` `Array(n).fill(1);` ` ` `for` `(let i = 0; i < n; i++)` ` ` `{` ` ` `for` `(let 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` ` ` `return` `dp.sort((a, b) => b - a)[0];` `}` `// Driver Code` `let arr = [1, 3, 2, 4, 5];` `let D = 2;` `document.write(longestSubsequence(arr, D));` `// This code is contributed by gfgking.` `</script>` |

**Output:**

3

**Time Complexity:** O(N^{2})**Auxiliary Space:** O(N)