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. ` `?> ` |

Output:

6

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

**Auxiliary Space: ** O(n)

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