Given an array of n size, the task is to find the longest subsequence such that difference between adjacents is one.

Examples:

Input :arr[] = {10, 9, 4, 5, 4, 8, 6}Output :3 As longest subsequences with difference 1 are, "10, 9, 8", "4, 5, 4" and "4, 5, 6"Input :arr[] = {1, 2, 3, 2, 3, 7, 2, 1}Output :7 As longest consecutive sequence is "1, 2, 3, 2, 3, 2, 1"

This problem is based upon the concept of Longest Increasing Subsequence Problem.

Let arr[0..n-1] be the input array and dp[i] be the length of the longest subsequence (with differences one) ending at index i such that arr[i] is the last element of the subsequence. Then, dp[i] can be recursively written as: dp[i] = 1 + max(dp[j]) where 0 < j < i and [arr[j] = arr[i] -1 or arr[j] = arr[i] + 1] dp[i] = 1, if no such j exists. To find the result for a given array, we need to return max(dp[i]) where 0 < i < n.

Following is a Dynamic Programming based implementation. It follows the recursive structure discussed above.

## C++

`// C++ program to find the longest subsequence such` `// the difference between adjacent elements of the` `// subsequence is one.` `#include<bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the length of longest subsequence` `int` `longestSubseqWithDiffOne(` `int` `arr[], ` `int` `n)` `{` ` ` `// Initialize the dp[] array with 1 as a` ` ` `// single element will be of 1 length` ` ` `int` `dp[n];` ` ` `for` `(` `int` `i = 0; i< n; i++)` ` ` `dp[i] = 1;` ` ` `// Start traversing the given array` ` ` `for` `(` `int` `i=1; i<n; i++)` ` ` `{` ` ` `// Compare with all the previous elements` ` ` `for` `(` `int` `j=0; j<i; j++)` ` ` `{` ` ` `// If the element is consecutive then` ` ` `// consider this subsequence and update` ` ` `// dp[i] if required.` ` ` `if` `((arr[i] == arr[j]+1) ||` ` ` `(arr[i] == arr[j]-1))` ` ` `dp[i] = max(dp[i], dp[j]+1);` ` ` `}` ` ` `}` ` ` `// Longest length will be the maximum value` ` ` `// of dp array.` ` ` `int` `result = 1;` ` ` `for` `(` `int` `i = 0 ; i < n ; i++)` ` ` `if` `(result < dp[i])` ` ` `result = dp[i];` ` ` `return` `result;` `}` `// Driver code` `int` `main()` `{` ` ` `// Longest subsequence with one difference is` ` ` `// {1, 2, 3, 4, 3, 2}` ` ` `int` `arr[] = {1, 2, 3, 4, 5, 3, 2};` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]);` ` ` `cout << longestSubseqWithDiffOne(arr, n);` ` ` `return` `0;` `}` |

## Java

`// Java program to find the longest subsequence` `// such that the difference between adjacent` `// elements of the subsequence is one.` `import` `java.io.*;` `class` `GFG {` ` ` ` ` `// Function to find the length of longest` ` ` `// subsequence` ` ` `static` `int` `longestSubseqWithDiffOne(` `int` `arr[],` ` ` `int` `n)` ` ` `{` ` ` `// Initialize the dp[] array with 1 as a` ` ` `// single element will be of 1 length` ` ` `int` `dp[] = ` `new` `int` `[n];` ` ` `for` `(` `int` `i = ` `0` `; i< n; i++)` ` ` `dp[i] = ` `1` `;` ` ` `// Start traversing the given array` ` ` `for` `(` `int` `i = ` `1` `; i < n; i++)` ` ` `{` ` ` `// Compare with all the previous` ` ` `// elements` ` ` `for` `(` `int` `j = ` `0` `; j < i; j++)` ` ` `{` ` ` `// If the element is consecutive` ` ` `// then consider this subsequence` ` ` `// and update dp[i] if required.` ` ` `if` `((arr[i] == arr[j] + ` `1` `) ||` ` ` `(arr[i] == arr[j] - ` `1` `))` ` ` `dp[i] = Math.max(dp[i], dp[j]+` `1` `);` ` ` `}` ` ` `}` ` ` `// Longest length will be the maximum` ` ` `// value of dp array.` ` ` `int` `result = ` `1` `;` ` ` `for` `(` `int` `i = ` `0` `; i < n ; i++)` ` ` `if` `(result < dp[i])` ` ` `result = dp[i];` ` ` `return` `result;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `// Longest subsequence with one` ` ` `// difference is` ` ` `// {1, 2, 3, 4, 3, 2}` ` ` `int` `arr[] = {` `1` `, ` `2` `, ` `3` `, ` `4` `, ` `5` `, ` `3` `, ` `2` `};` ` ` `int` `n = arr.length;` ` ` `System.out.println(longestSubseqWithDiffOne(` ` ` `arr, n));` ` ` `}` `}` `// This code is contributed by Prerna Saini` |

## Python

`# Function to find the length of longest subsequence` `def` `longestSubseqWithDiffOne(arr, n):` ` ` `# Initialize the dp[] array with 1 as a` ` ` `# single element will be of 1 length` ` ` `dp ` `=` `[` `1` `for` `i ` `in` `range` `(n)]` ` ` `# Start traversing the given array` ` ` `for` `i ` `in` `range` `(n):` ` ` `# Compare with all the previous elements` ` ` `for` `j ` `in` `range` `(i):` ` ` `# If the element is consecutive then` ` ` `# consider this subsequence and update` ` ` `# dp[i] if required.` ` ` `if` `((arr[i] ` `=` `=` `arr[j]` `+` `1` `) ` `or` `(arr[i] ` `=` `=` `arr[j]` `-` `1` `)):` ` ` `dp[i] ` `=` `max` `(dp[i], dp[j]` `+` `1` `)` ` ` `# Longest length will be the maximum value` ` ` `# of dp array.` ` ` `result ` `=` `1` ` ` `for` `i ` `in` `range` `(n):` ` ` `if` `(result < dp[i]):` ` ` `result ` `=` `dp[i]` ` ` ` ` `return` `result` `# Driver code` `arr ` `=` `[` `1` `, ` `2` `, ` `3` `, ` `4` `, ` `5` `, ` `3` `, ` `2` `]` `# Longest subsequence with one difference is` `# {1, 2, 3, 4, 3, 2}` `n ` `=` `len` `(arr)` `print` `longestSubseqWithDiffOne(arr, n)` `# This code is contributed by Afzal Ansari` |

## C#

`// C# program to find the longest subsequence` `// such that the difference between adjacent` `// elements of the subsequence is one.` `using` `System;` `class` `GFG {` ` ` ` ` `// Function to find the length of longest` ` ` `// subsequence` ` ` `static` `int` `longestSubseqWithDiffOne(` `int` `[]arr,` ` ` `int` `n)` ` ` `{` ` ` ` ` `// Initialize the dp[] array with 1 as a` ` ` `// single element will be of 1 length` ` ` `int` `[]dp = ` `new` `int` `[n];` ` ` ` ` `for` `(` `int` `i = 0; i< n; i++)` ` ` `dp[i] = 1;` ` ` `// Start traversing the given array` ` ` `for` `(` `int` `i = 1; i < n; i++)` ` ` `{` ` ` ` ` `// Compare with all the previous` ` ` `// elements` ` ` `for` `(` `int` `j = 0; j < i; j++)` ` ` `{` ` ` `// If the element is consecutive` ` ` `// then consider this subsequence` ` ` `// and update dp[i] if required.` ` ` `if` `((arr[i] == arr[j] + 1) ||` ` ` `(arr[i] == arr[j] - 1))` ` ` `dp[i] = Math.Max(dp[i], dp[j]+1);` ` ` `}` ` ` `}` ` ` `// Longest length will be the maximum` ` ` `// value of dp array.` ` ` `int` `result = 1;` ` ` `for` `(` `int` `i = 0 ; i < n ; i++)` ` ` `if` `(result < dp[i])` ` ` `result = dp[i];` ` ` ` ` `return` `result;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` ` ` `// Longest subsequence with one` ` ` `// difference is` ` ` `// {1, 2, 3, 4, 3, 2}` ` ` `int` `[]arr = {1, 2, 3, 4, 5, 3, 2};` ` ` `int` `n = arr.Length;` ` ` ` ` `Console.Write(` ` ` `longestSubseqWithDiffOne(arr, n));` ` ` `}` `}` `// This code is contributed by nitin mittal.` |

## PHP

`<?php` `// PHP program to find the longest` `// subsequence such the difference` `// between adjacent elements of the` `// subsequence is one.` `// Function to find the length of` `// longest subsequence` `function` `longestSubseqWithDiffOne(` `$arr` `, ` `$n` `)` `{` ` ` ` ` `// Initialize the dp[]` ` ` `// array with 1 as a` ` ` `// single element will` ` ` `// be of 1 length` ` ` `$dp` `[` `$n` `] = 0;` ` ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `$dp` `[` `$i` `] = 1;` ` ` `// Start traversing the` ` ` `// given array` ` ` `for` `(` `$i` `= 1; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `{` ` ` ` ` `// Compare with all the` ` ` `// previous elements` ` ` `for` `(` `$j` `= 0; ` `$j` `< ` `$i` `; ` `$j` `++)` ` ` `{` ` ` ` ` `// If the element is` ` ` `// consecutive then` ` ` `// consider this` ` ` `// subsequence and` ` ` `// update dp[i] if` ` ` `// required.` ` ` `if` `((` `$arr` `[` `$i` `] == ` `$arr` `[` `$j` `] + 1) ||` ` ` `(` `$arr` `[` `$i` `] == ` `$arr` `[` `$j` `] - 1))` ` ` `$dp` `[` `$i` `] = max(` `$dp` `[` `$i` `],` ` ` `$dp` `[` `$j` `] + 1);` ` ` `}` ` ` `}` ` ` `// Longest length will be` ` ` `// the maximum value` ` ` `// of dp array.` ` ` `$result` `= 1;` ` ` `for` `(` `$i` `= 0 ; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `if` `(` `$result` `< ` `$dp` `[` `$i` `])` ` ` `$result` `= ` `$dp` `[` `$i` `];` ` ` `return` `$result` `;` `}` ` ` `// Driver code` ` ` `// Longest subsequence with` ` ` `// one difference is` ` ` `// {1, 2, 3, 4, 3, 2}` ` ` `$arr` `= ` `array` `(1, 2, 3, 4, 5, 3, 2);` ` ` `$n` `= sizeof(` `$arr` `);` ` ` `echo` `longestSubseqWithDiffOne(` `$arr` `, ` `$n` `);` ` ` `// This code is contributed by nitin mittal.` `?>` |

## Javascript

`<script>` `// JavaScript program to find the` `// longest subsequence such that the` `// difference between adjacent elements` `// of the subsequence is one.` `// Function to find the length of longest` `// subsequence` `function` `longestSubseqWithDiffOne(arr, n)` `{` ` ` ` ` `// Initialize the dp[] array with 1 as a` ` ` `// single element will be of 1 length` ` ` `let dp = [];` ` ` `for` `(let i = 0; i < n; i++)` ` ` `dp[i] = 1;` ` ` ` ` `// Start traversing the given array` ` ` `for` `(let i = 1; i < n; i++)` ` ` `{` ` ` ` ` `// Compare with all the previous` ` ` `// elements` ` ` `for` `(let j = 0; j < i; j++)` ` ` `{` ` ` ` ` `// If the element is consecutive` ` ` `// then consider this subsequence` ` ` `// and update dp[i] if required.` ` ` `if` `((arr[i] == arr[j] + 1) ||` ` ` `(arr[i] == arr[j] - 1))` ` ` ` ` `dp[i] = Math.max(dp[i], dp[j] + 1);` ` ` `}` ` ` `}` ` ` ` ` `// Longest length will be the maximum` ` ` `// value of dp array.` ` ` `let result = 1;` ` ` `for` `(let i = 0; i < n; i++)` ` ` `if` `(result < dp[i])` ` ` `result = dp[i];` ` ` ` ` `return` `result;` `}` `// Driver Code` `// Longest subsequence with one` `// difference is` `// {1, 2, 3, 4, 3, 2}` `let arr = [1, 2, 3, 4, 5, 3, 2];` `let n = arr.length;` `document.write(longestSubseqWithDiffOne(arr, n));` `// This code is contributed by souravghosh0416` `</script>` |

**Output: **

6

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

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