Given an array **arr[]** consisting of **N** integers, the task is to find the length of the longest non-decreasing subsequence such that the difference between adjacent elements is at most **1**.

**Examples:**

Input:arr[] = {8, 5, 4, 8, 4}Output:3Explanation:{4, 4, 5}, {8, 8} are the two such non-decreasing subsequences of length 2 and 3 respectively. Therefore, the length of the longest of the two subsequences is 3.

Input:arr[] = {4, 13, 2, 3}Output:3Explanation:{2, 3, 4}, {13} are the two such non-decreasing subsequences of length 3 and 1 respectively. Therefore, the length of the longest of the two subsequences is 3.

**Approach**: Follow the steps below to solve the problem:

- Sort the array arr[] in increasing order.
- Initialize a variable, say
**maxLen = 1,**to store the maximum possible length of a subsequence. Initialize another variable, say**len = 1**to store the current length for each subsequence. - Traverse the array
**arr[]**with a pointer**i**and for each element:- Check if
**abs(arr[i] – arr[i – 1]) ≤ 1**. If found to be true, then increment**len**by**1**. Update**maxLen = max(maxLen, len)**. - Otherwise, set
**len = 1**i.e. start a new subsequence.

- Check if
- Print the value of
**maxLen**as the final answer.

Below is the implementation of the above approach:

## C++

`// C++ program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the longest non-decreasing` `// subsequence with difference between` `// adjacent elements exactly equal to 1` `void` `longestSequence(` `int` `arr[], ` `int` `N)` `{` ` ` `// Base case` ` ` `if` `(N == 0) {` ` ` `cout << 0;` ` ` `return` `;` ` ` `}` ` ` `// Sort the array in ascending order` ` ` `sort(arr, arr+N);` ` ` `// Stores the maximum length` ` ` `int` `maxLen = 1;` ` ` `int` `len = 1;` ` ` `// Traverse the array` ` ` `for` `(` `int` `i = 1; i < N; i++) {` ` ` `// If difference between current` ` ` `// pair of adjacent elements is 1 or 0` ` ` `if` `(arr[i] == arr[i - 1]` ` ` `|| arr[i] == arr[i - 1] + 1) {` ` ` `len++;` ` ` `// Extend the current sequence` ` ` `// Update len and max_len` ` ` `maxLen = max(maxLen, len);` ` ` `}` ` ` `else` `{` ` ` `// Otherwise, start a new subsequence` ` ` `len = 1;` ` ` `}` ` ` `}` ` ` `// Print the maximum length` ` ` `cout << maxLen;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given array` ` ` `int` `arr[] = { 8, 5, 4, 8, 4 };` ` ` `// Size of the array` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `// Function call to find the longest` ` ` `// subsequence` ` ` `longestSequence(arr, N);` ` ` `return` `0;` `}` `// This code is contributed by code_hunt.` |

## Java

`// Java program for the above approach` `import` `java.io.*;` `import` `java.util.*;` `class` `GFG {` ` ` `// Function to find the longest non-decreasing` ` ` `// subsequence with difference between` ` ` `// adjacent elements exactly equal to 1` ` ` `static` `void` `longestSequence(` `int` `arr[], ` `int` `N)` ` ` `{` ` ` `// Base case` ` ` `if` `(N == ` `0` `) {` ` ` `System.out.println(` `0` `);` ` ` `return` `;` ` ` `}` ` ` `// Sort the array in ascending order` ` ` `Arrays.sort(arr);` ` ` `// Stores the maximum length` ` ` `int` `maxLen = ` `1` `;` ` ` `int` `len = ` `1` `;` ` ` `// Traverse the array` ` ` `for` `(` `int` `i = ` `1` `; i < N; i++) {` ` ` `// If difference between current` ` ` `// pair of adjacent elements is 1 or 0` ` ` `if` `(arr[i] == arr[i - ` `1` `]` ` ` `|| arr[i] == arr[i - ` `1` `] + ` `1` `) {` ` ` `len++;` ` ` `// Extend the current sequence` ` ` `// Update len and max_len` ` ` `maxLen = Math.max(maxLen, len);` ` ` `}` ` ` `else` `{` ` ` `// Otherwise, start a new subsequence` ` ` `len = ` `1` `;` ` ` `}` ` ` `}` ` ` `// Print the maximum length` ` ` `System.out.println(maxLen);` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `// Given array` ` ` `int` `arr[] = { ` `8` `, ` `5` `, ` `4` `, ` `8` `, ` `4` `};` ` ` `// Size of the array` ` ` `int` `N = arr.length;` ` ` `// Function call to find the longest` ` ` `// subsequence` ` ` `longestSequence(arr, N);` ` ` `}` `}` |

## Python3

`# Python program for the above approach` `# Function to find the longest non-decreasing` `# subsequence with difference between` `# adjacent elements exactly equal to 1` `def` `longestSequence(arr, N):` ` ` ` ` `# Base case` ` ` `if` `(N ` `=` `=` `0` `):` ` ` `print` `(` `0` `);` ` ` `return` `;` ` ` `# Sort the array in ascending order` ` ` `arr.sort();` ` ` `# Stores the maximum length` ` ` `maxLen ` `=` `1` `;` ` ` `len` `=` `1` `;` ` ` `# Traverse the array` ` ` `for` `i ` `in` `range` `(` `1` `,N):` ` ` `# If difference between current` ` ` `# pair of adjacent elements is 1 or 0` ` ` `if` `(arr[i] ` `=` `=` `arr[i ` `-` `1` `] ` `or` `arr[i] ` `=` `=` `arr[i ` `-` `1` `] ` `+` `1` `):` ` ` `len` `+` `=` `1` `;` ` ` `# Extend the current sequence` ` ` `# Update len and max_len` ` ` `maxLen ` `=` `max` `(maxLen, ` `len` `);` ` ` `else` `:` ` ` `# Otherwise, start a new subsequence` ` ` `len` `=` `1` `;` ` ` `# Prthe maximum length` ` ` `print` `(maxLen);` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `# Given array` ` ` `arr ` `=` `[` `8` `, ` `5` `, ` `4` `, ` `8` `, ` `4` `];` ` ` `# Size of the array` ` ` `N ` `=` `len` `(arr);` ` ` `# Function call to find the longest` ` ` `# subsequence` ` ` `longestSequence(arr, N);` ` ` `# This code is contributed by 29AjayKumar` |

## C#

`// C# program for the above approach` `using` `System;` `public` `class` `GFG` `{` ` ` `// Function to find the longest non-decreasing` ` ` `// subsequence with difference between` ` ` `// adjacent elements exactly equal to 1` ` ` `static` `void` `longestSequence(` `int` `[]arr, ` `int` `N)` ` ` `{` ` ` `// Base case` ` ` `if` `(N == 0)` ` ` `{` ` ` `Console.WriteLine(0);` ` ` `return` `;` ` ` `}` ` ` `// Sort the array in ascending order` ` ` `Array.Sort(arr);` ` ` `// Stores the maximum length` ` ` `int` `maxLen = 1;` ` ` `int` `len = 1;` ` ` `// Traverse the array` ` ` `for` `(` `int` `i = 1; i < N; i++) {` ` ` `// If difference between current` ` ` `// pair of adjacent elements is 1 or 0` ` ` `if` `(arr[i] == arr[i - 1]` ` ` `|| arr[i] == arr[i - 1] + 1) {` ` ` `len++;` ` ` `// Extend the current sequence` ` ` `// Update len and max_len` ` ` `maxLen = Math.Max(maxLen, len);` ` ` `}` ` ` `else` `{` ` ` `// Otherwise, start a new subsequence` ` ` `len = 1;` ` ` `}` ` ` `}` ` ` `// Print the maximum length` ` ` `Console.WriteLine(maxLen);` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main(` `string` `[] args)` ` ` `{` ` ` `// Given array` ` ` `int` `[]arr = { 8, 5, 4, 8, 4 };` ` ` `// Size of the array` ` ` `int` `N = arr.Length;` ` ` `// Function call to find the longest` ` ` `// subsequence` ` ` `longestSequence(arr, N);` ` ` `}` `}` `// This code is contributed by AnkThon` |

## Javascript

`<script>` `// Javascript program to implement` `// the above approach` `// Function to find the longest non-decreasing` `// subsequence with difference between` `// adjacent elements exactly equal to 1` `function` `longestSequence(arr, N)` `{` ` ` `// Base case` ` ` `if` `(N == 0) {` ` ` `document.write(0);` ` ` `return` `;` ` ` `}` ` ` `// Sort the array in ascending order` ` ` `arr.sort();` ` ` `// Stores the maximum length` ` ` `var` `maxLen = 1;` ` ` `var` `len = 1;` ` ` `var` `i;` ` ` `// Traverse the array` ` ` `for` `(i = 1; i < N; i++) {` ` ` `// If difference between current` ` ` `// pair of adjacent elements is 1 or 0` ` ` `if` `(arr[i] == arr[i - 1]` ` ` `|| arr[i] == arr[i - 1] + 1) {` ` ` `len++;` ` ` `// Extend the current sequence` ` ` `// Update len and max_len` ` ` `maxLen = Math.max(maxLen, len);` ` ` `}` ` ` `else` `{` ` ` `// Otherwise, start a new subsequence` ` ` `len = 1;` ` ` `}` ` ` `}` ` ` `// Print the maximum length` ` ` `document.write(maxLen);` `}` `// Driver Code` ` ` `// Given array` ` ` `var` `arr = [8, 5, 4, 8, 4];` ` ` `// Size of the array` ` ` `var` `N = arr.length;` ` ` `// Function call to find the longest` ` ` `// subsequence` ` ` `longestSequence(arr, N);` `</script>` |

**Output:**

3

**Time Complexity:** O(N * logN)**Auxiliary Space:** O(1)

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