Given an array arr[] of size N and an integer K, the task is to find the longest subsequence such that the difference between adjacents is K.
Example:
Input: arr[]={1, 2, 3, 4, 5, 3, 2}, K=1
Output: 6
Explanation: The longest subsequence with the difference between the adjacent elements as 1 is: {1, 2, 3, 4, 3, 2}Input: arr[]={1, 2, 3, 2, 3, 7, 2, 1}, K=2
Output: 3
Approach: The given problem can be solved using Dynamic Programming. The idea is to store the length of the longest subsequence having a difference K between adjacents ending after including the current element. Create an unordered map mp where mp[i] represents the maximum length of subsequence which includes integer i.
So, the relation to get the maximum length subsequence can be written as:
mp[i] = 1 + max(mp[i – K], mp[i + K])
The maximum integer stores in the map mp is the required answer. Below is the implementation of the above approach:
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the longest// subsequence such that difference// between adjacent elements is Kint longestSubsequence(vector<int>& arr, int K){ // Stores length of longest // subsequence in a map unordered_map<int, int> mp; // Variable to store the maximum // length of subsequence int mx = 1; // Loop to iterate through the // given array for (auto x : arr) { mp[x] = 1; // If (x - K) exists if (mp.count(x - K)) { mp[x] = 1 + mp[x - K]; } // if (x + K) exists if (mp.count(x + K)) { mp[x] = max(mp[x], 1 + mp[x + K]); } mx = max(mx, mp[x]); } // Return Answer return mx;}// Driver Codeint main(){ vector<int> arr = { 1, 2, 3, 4, 5, 3, 2 }; int K = 1; cout << longestSubsequence(arr, K);} |
Java
// Java code for the above approachimport java.util.HashMap;class GFG { // Function to find the longest // subsequence such that difference // between adjacent elements is K public static int longestSubsequence(int[] arr, int K) { // Stores length of longest // subsequence in a map HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>(); // Variable to store the maximum // length of subsequence int mx = 1; // Loop to iterate through the // given array for (int x : arr) { mp.put(x, 1); // If (x - K) exists if (mp.containsKey(x - K)) { mp.put(x, 1 + mp.get(x - K)); } // If (x + K) exists if (mp.containsKey(x + K)) { mp.put(x, Math.max(mp.get(x), 1 + mp.get(x + K))); } mx = Math.max(mx, mp.get(x)); } // Return Answer return mx; } // Driver Code public static void main(String args[]) { int[] arr = { 1, 2, 3, 4, 5, 3, 2 }; int K = 1; System.out.print(longestSubsequence(arr, K)); }}// This code is contributed by gfgking |
Python3
# python code for the above approach# Function to find the longest# subsequence such that difference# between adjacent elements is Kdef longestSubsequence(arr, K): # Stores length of longest # subsequence in a map mp = {} # Variable to store the maximum # length of subsequence mx = 1 # Loop to iterate through the # given array for x in arr: mp[x] = 1 # If (x - K) exists if ((x - K) in mp): mp[x] = 1 + mp[x - K] # if (x + K) exists if ((x+K) in mp): mp[x] = max(mp[x], 1 + mp[x + K]) mx = max(mx, mp[x]) # Return Answer return mx# Driver Codeif __name__ == "__main__": arr = [1, 2, 3, 4, 5, 3, 2] K = 1 print(longestSubsequence(arr, K))# This code is contributed by rakeshsahni |
C#
// C# code for the above approachusing System;using System.Collections.Generic;class GFG{ // Function to find the longest// subsequence such that difference// between adjacent elements is Kstatic int longestSubsequence(List<int> arr, int K){ // Stores length of longest // subsequence in a map Dictionary<int, int> mp = new Dictionary<int, int>(); // Variable to store the maximum // length of subsequence int mx = 1; // Loop to iterate through the // given array foreach(int x in arr) { mp[x] = 1; // If (x - K) exists if (mp.ContainsKey(x - K)) { mp[x] = 1 + mp[x - K]; } // If (x + K) exists if (mp.ContainsKey(x + K)) { mp[x] = Math.Max(mp[x], 1 + mp[x + K]); } mx = Math.Max(mx, mp[x]); } // Return Answer return mx;}// Driver Codepublic static void Main(){ List<int> arr = new List<int>(){ 1, 2, 3, 4, 5, 3, 2 }; int K = 1; Console.Write(longestSubsequence(arr, K));}}// This code is contributed by ukasp |
Javascript
<script> // JavaScript code for the above approach // Function to find the longest // subsequence such that difference // between adjacent elements is K function longestSubsequence(arr, K) { // Stores length of longest // subsequence in a map let mp = new Map(); // Variable to store the maximum // length of subsequence let mx = 1; // Loop to iterate through the // given array for (let x of arr) { mp.set(x, 1) // If (x - K) exists if (mp.has(x - K)) { mp.set(x, 1 + mp.get(x - K)); } // if (x + K) exists if (mp.has(x + K)) { mp.set(x, 1 + mp.get(x + K)); } mx = Math.max(mx, mp.get(x)); } // Return Answer return mx; } // Driver Code let arr = [1, 2, 3, 4, 5, 3, 2]; let K = 1; document.write(longestSubsequence(arr, K));// This code is contributed by Potta Lokesh </script> |
Output
6
Time Complexity: O(N)
Auxiliary Space: O(N)
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