Given an array of n integers. The problem is to find maximum length of the subsequence with difference between adjacent elements in the subsequence as either 0 or 1. Time Complexity of O(n) is required.
Examples:
Input : arr[] = {2, 5, 6, 3, 7, 6, 5, 8}
Output : 5
The subsequence is {5, 6, 7, 6, 5}.
Input : arr[] = {-2, -1, 5, -1, 4, 0, 3}
Output : 4
The subsequence is {-2, -1, -1, 0}.
Method 1: Previously an approach having time complexity of O(n2) have been discussed in this post.
Method 2 (Efficient Approach):
The idea is to create a hash map having tuples in the form (ele, len), where len denotes the length of the longest subsequence ending with the element ele. Now, for each element arr[i] we can find the length of the values arr[i]-1, arr[i] and arr[i]+1 in the hash table and consider the maximum among them. Let this maximum value be max. Now, the length of longest subsequence ending with arr[i] would be max+1. Update this length along with the element arr[i] in the hash table. Finally, the element having the maximum length in the hash table gives the maximum length subsequence.
Implementation:
C++
// C++ implementation to find maximum length// subsequence with difference between adjacent // elements as either 0 or 1#include <bits/stdc++.h>using namespace std; // function to find maximum length subsequence // with difference between adjacent elements as// either 0 or 1int maxLenSub(int arr[], int n){ // hash table to map the array element with the // length of the longest subsequence of which // it is a part of and is the last element of // that subsequence unordered_map<int, int> um; // to store the maximum length subsequence int maxLen = 0; // traverse the array elements for (int i=0; i<n; i++) { // initialize current length // for element arr[i] as 0 int len = 0; // if 'arr[i]-1' is in 'um' and its length of // subsequence is greater than 'len' if (um.find(arr[i]-1) != um.end() && len < um[arr[i]-1]) len = um[arr[i]-1]; // if 'arr[i]' is in 'um' and its length of // subsequence is greater than 'len' if (um.find(arr[i]) != um.end() && len < um[arr[i]]) len = um[arr[i]]; // if 'arr[i]+1' is in 'um' and its length of // subsequence is greater than 'len' if (um.find(arr[i]+1) != um.end() && len < um[arr[i]+1]) len = um[arr[i]+1]; // update arr[i] subsequence length in 'um' um[arr[i]] = len + 1; // update maximum length if (maxLen < um[arr[i]]) maxLen = um[arr[i]]; } // required maximum length subsequence return maxLen; } // Driver program to test aboveint main(){ int arr[] = {2, 5, 6, 3, 7, 6, 5, 8}; int n = sizeof(arr) / sizeof(arr[0]); cout << "Maximum length subsequence = " << maxLenSub(arr, n); return 0;} |
Java
// Java implementation to find maximum length// subsequence with difference between adjacent // elements as either 0 or 1import java.util.HashMap;class GFG{ // function to find maximum length subsequence // with difference between adjacent elements as // either 0 or 1 public static int maxLengthSub(int[] arr) { // to store the maximum length subsequence int max_val = 0; int start = 0; // hash table to map the array element with the // length of the longest subsequence of which // it is a part of and is the last element of // that subsequence HashMap<Integer, Integer> map = new HashMap<>(); // traverse the array elements for (int i = 0; i < arr.length; i++) { // initialize current length // for element arr[i] as 0 int temp = 0; if (map.containsKey(arr[i] - 1)) { temp = map.get(arr[i] - 1); } if (map.containsKey(arr[i])) { temp = Math.max(temp, map.get(arr[i])); } if (map.containsKey(arr[i] + 1)) { temp = Math.max(temp, map.get(arr[i] + 1)); } temp++; // update maximum length if (temp > max_val) { max_val = temp; } map.put(arr[i], temp); } // required maximum length subsequence return max_val; } // Driver Code public static void main(String[] args) { int arr[] = {2, 5, 6, 3, 7, 6, 5, 8}; System.out.println(maxLengthSub(arr)); }}// This code is contributed // by tushar jajodia |
Python3
# Python3 implementation to find maximum # length subsequence with difference between # adjacent elements as either 0 or 1 from collections import defaultdict# Function to find maximum length subsequence with # difference between adjacent elements as either 0 or 1 def maxLenSub(arr, n): # hash table to map the array element with the # length of the longest subsequence of which it is a # part of and is the last element of that subsequence um = defaultdict(lambda:0) # to store the maximum length subsequence maxLen = 0 # traverse the array elements for i in range(0, n): # initialize current length # for element arr[i] as 0 length = 0 # if 'arr[i]-1' is in 'um' and its length of # subsequence is greater than 'len' if (arr[i]-1) in um and length < um[arr[i]-1]: length = um[arr[i]-1] # if 'arr[i]' is in 'um' and its length of # subsequence is greater than 'len' if arr[i] in um and length < um[arr[i]]: length = um[arr[i]] # if 'arr[i]+1' is in 'um' and its length of # subsequence is greater than 'len' if (arr[i]+1) in um and length < um[arr[i]+1]: length = um[arr[i]+1] # update arr[i] subsequence length in 'um' um[arr[i]] = length + 1 # update maximum length if maxLen < um[arr[i]]: maxLen = um[arr[i]] # required maximum length subsequence return maxLen# Driver program to test above if __name__ == "__main__": arr = [2, 5, 6, 3, 7, 6, 5, 8] n = len(arr) print("Maximum length subsequence =", maxLenSub(arr, n)) # This code is contributed by Rituraj Jain |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// function to find maximum length subsequence // with difference between adjacent elements as// either 0 or 1public static int maxLenSub(int[] arr, int n){ // hash table to map the array element with the // length of the longest subsequence of which // it is a part of and is the last element of // that subsequence Dictionary<int, int> um = new Dictionary<int, int>(); // to store the maximum length subsequence int maxLen = 0; // traverse the array elements for (int i=0; i<n; i++) { // initialize current length // for element arr[i] as 0 int len = 0; // if 'arr[i]-1' is in 'um' and its length of // subsequence is greater than 'len' if (um.ContainsKey(arr[i] - 1) && len < um[arr[i]-1]) len = um[arr[i]-1]; // if 'arr[i]' is in 'um' and its length of // subsequence is greater than 'len' if (um.ContainsKey(arr[i]) && len < um[arr[i]]) len = um[arr[i]]; // if 'arr[i]+1' is in 'um' and its length of // subsequence is greater than 'len' if (um.ContainsKey(arr[i]+1) && len < um[arr[i]+1]) len = um[arr[i]+1]; // update arr[i] subsequence length in 'um' um[arr[i]] = len + 1; // update maximum length if (maxLen < um[arr[i]]) maxLen = um[arr[i]]; } // required maximum length subsequence return maxLen; }// Driver Codepublic static void Main(){ int[] arr = {2, 5, 6, 3, 7, 6, 5, 8}; int n = arr.Length; Console.WriteLine("Maximum length subsequence = " + maxLenSub(arr, n));}}// This code is contributed by sanjoy_62. |
Javascript
<script>// Javascript implementation to find maximum length// subsequence with difference between adjacent // elements as either 0 or 1 // function to find maximum length subsequence // with difference between adjacent elements as// either 0 or 1function maxLenSub(arr, n){ // hash table to map the array element with the // length of the longest subsequence of which // it is a part of and is the last element of // that subsequence var um = new Map(); // to store the maximum length subsequence var maxLen = 0; // traverse the array elements for (var i=0; i<n; i++) { // initialize current length // for element arr[i] as 0 var len = 0; // if 'arr[i]-1' is in 'um' and its length of // subsequence is greater than 'len' if (um.has(arr[i]-1) && len < um.get(arr[i]-1)) len = um.get(arr[i]-1); // if 'arr[i]' is in 'um' and its length of // subsequence is greater than 'len' if (um.has(arr[i]) && len < um.get(arr[i])) len = um.get(arr[i]); // if 'arr[i]+1' is in 'um' and its length of // subsequence is greater than 'len' if (um.has(arr[i]+1) && len < um.get(arr[i]+1)) len = um.get(arr[i]+1); // update arr[i] subsequence length in 'um' um.set(arr[i], len + 1); // update maximum length if (maxLen < um.get(arr[i])) maxLen = um.get(arr[i]); } // required maximum length subsequence return maxLen; } // Driver program to test abovevar arr = [2, 5, 6, 3, 7, 6, 5, 8];var n = arr.length;document.write( "Maximum length subsequence = " + maxLenSub(arr, n));// This code is contributed by itsok.</script> |
Output
Maximum length subsequence = 5
Time Complexity: O(n)
Auxiliary Space: O(n)
Thanks to Neeraj for suggesting the above solution in the comments of this post.
This article is contributed by Ayush Jauhari. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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