Longest subsequence such that difference between adjacents is one - GeeksforGeeks

25 Mar, 2022
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++
Java
Python3
C#
PHP
Javascript
// 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 subsequenceint 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 codeint 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 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
# Function to find the length of longest subsequencedef 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 codearr = [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# 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 program to find the longest// subsequence such the difference// between adjacent elements of the// subsequence is one. // Function to find the length of// longest subsequencefunction 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.?>
<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// subsequencefunction 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>
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Time Complexity: O(n2) Auxiliary Space: O(n)
Efficient Approach 
C++
Java
Python3
#include<bits/stdc++.h>using namespace std;int longestSubsequence(int n, int arr[])    {        if(n==1)            return 1;        unordered_map<int,int> mapp;        int res = 1;        for(int i=0;i<n;i++){            if(mapp.count(arr[i]+1) >0 || mapp.count(arr[i]-1)>0){                mapp[arr[i]]=1+max(mapp[arr[i]+1],mapp[arr[i]-1]);            }            else                mapp[arr[i]]=1;            res = max(res, mapp[arr[i]]);        }        return res;          //This code is contributed by Akansha Mittal    }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 << longestSubsequence(n, arr);    return 0;}
import java.lang.Math;import java.util.*; class GFG {     static int longestSubsequence(int n, int arr[])    {        if (n == 1)            return 1;        Integer dp[] = new Integer[n];        HashMap<Integer, Integer> mapp = new HashMap<>();        dp[0] = 1;        mapp.put(arr[0], 0);        for (int i = 1; i < n; i++) {            if (Math.abs(arr[i] - arr[i - 1]) == 1)                dp[i] = dp[i - 1] + 1;            else {                if (mapp.containsKey(arr[i] + 1)                    || mapp.containsKey(arr[i] - 1)) {                    dp[i] = 1                            + Math.max(mapp.getOrDefault(                                           arr[i] + 1, 0),                                       mapp.getOrDefault(                                           arr[i] - 1, 0));                }                else                    dp[i] = 1;            }            mapp.put(arr[i], dp[i]);        }        return Collections.max(Arrays.asList(dp));    }     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(longestSubsequence(n, arr));    }} // This code is contributed by rajsanghavi9.
def longestSubsequence(A, N):    L = [1]*N    hm = {}    for i in range(1,N):        if abs(A[i]-A[i-1]) == 1:            L[i] = 1 + L[i-1]        elif hm.get(A[i]+1,0) or hm.get(A[i]-1,0):            L[i] = 1+max(hm.get(A[i]+1,0), hm.get(A[i]-1,0))        hm[A[i]] = L[i]    return max(L)# Driver codeA =  [1, 2, 3, 4, 5, 3, 2]N = len(A)print(longestSubsequence(A, N))
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nitin mittal
souravghosh0416
rajsanghavi9
praveen7230
amartyaghoshgfg
aakankshamittal22
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Dynamic Programming
Dynamic Programming
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