Given N elements, write a program that prints the length of the longest increasing consecutive subsequence.
Examples:
Input : a[] = {3, 10, 3, 11, 4, 5, 6, 7, 8, 12}
Output : 6
Explanation: 3, 4, 5, 6, 7, 8 is the longest increasing subsequence whose adjacent element differs by one.Input : a[] = {6, 7, 8, 3, 4, 5, 9, 10}
Output : 5
Explanation: 6, 7, 8, 9, 10 is the longest increasing subsequence
Naive Approach: For every element, find the length of the subsequence starting from that particular element. Print the longest length of the subsequence thus formed:
C++
#include <bits/stdc++.h>using namespace std;int LongestSubsequence(int a[], int n){ int ans = 0; // Traverse every element to check if any // increasing subsequence starts from this index for(int i=0; i<n; i++) { // Initialize cnt variable as 1, which defines // the current length of the increasing subsequence int cnt = 1; for(int j=i+1; j<n; j++) if(a[j] == (a[i]+cnt)) cnt++; // Update the answer if the current length is // greater than already found length ans = max(ans, cnt); } return ans;}int main(){ int a[] = { 3, 10, 3, 11, 4, 5, 6, 7, 8, 12 }; int n = sizeof(a) / sizeof(a[0]); cout << LongestSubsequence(a, n); return 0;} |
Java
import java.util.Scanner;public class Main { public static int LongestSubsequence(int a[], int n) { int ans = 0; // Traverse every element to check if any // increasing subsequence starts from this index for(int i=0; i<n; i++) { // Initialize cnt variable as 1, which defines // the current length of the increasing subsequence int cnt = 1; for(int j=i+1; j<n; j++) if(a[j] == (a[i]+cnt)) cnt++; // Update the answer if the current length is // greater than already found length if(cnt > ans) ans = cnt; } return ans; } public static void main(String[] args) { int[] a = { 3, 10, 3, 11, 4, 5, 6, 7, 8, 12}; int n = a.length; System.out.println(LongestSubsequence(a, n)); }}// This code contributed by Ajax |
Python3
def longest_subsequence(a, n): ans = 0 # Traverse every element to check if any # increasing subsequence starts from this index for i in range(n): # Initialize cnt variable as 1, which defines # the current length of the increasing subsequence cnt = 1 for j in range(i + 1, n): if a[j] == (a[i] + cnt): cnt += 1 # Update the answer if the current length is # greater than the already found length ans = max(ans, cnt) return ansif __name__ == "__main__": a = [3, 10, 3, 11, 4, 5, 6, 7, 8, 12] n = len(a) print(longest_subsequence(a, n)) |
C#
using System;class GFG{ static int LongestSubsequence(int[] a, int n) { int ans = 0; // Traverse every element to check if any // increasing subsequence starts from this index for (int i = 0; i < n; i++) { // Initialize cnt variable as 1, which defines // current length of the increasing subsequence int cnt = 1; for (int j = i + 1; j < n; j++) { if (a[j] == (a[i] + cnt)) { cnt++; } // Update the answer if the current length is // greater than the already found length ans = Math.Max(ans, cnt); } } return ans; } static void Main() { int[] a = { 3, 10, 3, 11, 4, 5, 6, 7, 8, 12 }; int n = a.Length; Console.WriteLine(LongestSubsequence(a, n)); }} |
Javascript
function LongestSubsequence(a, n){ let ans = 0; // Traverse every element to check if any // increasing subsequence starts from this index for(let i=0; i<n; i++) { // Initialize cnt variable as 1, which defines // the current length of the increasing subsequence let cnt = 1; for(let j=i+1; j<n; j++) if(a[j] == (a[i]+cnt)) cnt++; // Update the answer if the current length is // greater than already found length ans = Math.max(ans, cnt); } return ans;}let a = [ 3, 10, 3, 11, 4, 5, 6, 7, 8, 12 ];let n = a.length;console.log(LongestSubsequence(a, n)); |
Output
6
Time Complexity: O(N2)
Auxiliary Space: O(1)
Dynamic Programming Approach: Let DP[i] store the length of the longest subsequence which ends with A[i]. For every A[i], if A[i]-1 is present in the array before i-th index, then A[i] will add to the increasing subsequence which has A[i]-1. Hence, DP[i] = DP[ index(A[i]-1) ] + 1. If A[i]-1 is not present in the array before i-th index, then DP[i]=1 since the A[i] element forms a subsequence which starts with A[i]. Hence, the relation for DP[i] is:
If A[i]-1 is present before i-th index:
- DP[i] = DP[ index(A[i]-1) ] + 1
else:
- DP[i] = 1
Given below is the illustration of the above approach:
C++
// CPP program to find length of the// longest increasing subsequence// whose adjacent element differ by 1#include <bits/stdc++.h>using namespace std;// function that returns the length of the// longest increasing subsequence// whose adjacent element differ by 1int longestSubsequence(int a[], int n){ // stores the index of elements unordered_map<int, int> mp; // stores the length of the longest // subsequence that ends with a[i] int dp[n]; memset(dp, 0, sizeof(dp)); int maximum = INT_MIN; // iterate for all element for (int i = 0; i < n; i++) { // if a[i]-1 is present before i-th index if (mp.find(a[i] - 1) != mp.end()) { // last index of a[i]-1 int lastIndex = mp[a[i] - 1] - 1; // relation dp[i] = 1 + dp[lastIndex]; } else dp[i] = 1; // stores the index as 1-index as we need to // check for occurrence, hence 0-th index // will not be possible to check mp[a[i]] = i + 1; // stores the longest length maximum = max(maximum, dp[i]); } return maximum;}// Driver Codeint main(){ int a[] = { 4, 3, 10, 3, 11, 4, 5, 6, 7, 8, 12 }; int n = sizeof(a) / sizeof(a[0]); cout << longestSubsequence(a, n); return 0;} |
Java
// Java program to find length of the// longest increasing subsequence// whose adjacent element differ by 1import java.util.*;class lics { static int LongIncrConseqSubseq(int arr[], int n) { // create hashmap to save latest consequent // number as "key" and its length as "value" HashMap<Integer, Integer> map = new HashMap<>(); // put first element as "key" and its length as "value" map.put(arr[0], 1); for (int i = 1; i < n; i++) { // check if last consequent of arr[i] exist or not if (map.containsKey(arr[i] - 1)) { // put the updated consequent number // and increment its value(length) map.put(arr[i], map.get(arr[i] - 1) + 1); // remove the last consequent number map.remove(arr[i] - 1); } // if there is no last consequent of // arr[i] then put arr[i] else { map.put(arr[i], 1); } } return Collections.max(map.values()); } // driver code public static void main(String args[]) { // Take input from user Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int arr[] = new int[n]; for (int i = 0; i < n; i++) arr[i] = sc.nextInt(); System.out.println(LongIncrConseqSubseq(arr, n)); }}// This code is contributed by CrappyDoctor |
Python3
# python program to find length of the # longest increasing subsequence # whose adjacent element differ by 1 from collections import defaultdictimport sys# function that returns the length of the # longest increasing subsequence # whose adjacent element differ by 1 def longestSubsequence(a, n): mp = defaultdict(lambda:0) # stores the length of the longest # subsequence that ends with a[i] dp = [0 for i in range(n)] maximum = -sys.maxsize # iterate for all element for i in range(n): # if a[i]-1 is present before i-th index if a[i] - 1 in mp: # last index of a[i]-1 lastIndex = mp[a[i] - 1] - 1 # relation dp[i] = 1 + dp[lastIndex] else: dp[i] = 1 # stores the index as 1-index as we need to # check for occurrence, hence 0-th index # will not be possible to check mp[a[i]] = i + 1 # stores the longest length maximum = max(maximum, dp[i]) return maximum# Driver Code a = [3, 10, 3, 11, 4, 5, 6, 7, 8, 12]n = len(a)print(longestSubsequence(a, n))# This code is contributed by Shrikant13 |
C#
// C# program to find length of the// longest increasing subsequence// whose adjacent element differ by 1using System;using System.Collections.Generic;class GFG{ static int longIncrConseqSubseq(int []arr, int n){ // Create hashmap to save // latest consequent number // as "key" and its length // as "value" Dictionary<int, int> map = new Dictionary<int, int>(); // Put first element as "key" // and its length as "value" map.Add(arr[0], 1); for (int i = 1; i < n; i++) { // Check if last consequent // of arr[i] exist or not if (map.ContainsKey(arr[i] - 1)) { // put the updated consequent number // and increment its value(length) map.Add(arr[i], map[arr[i] - 1] + 1); // Remove the last consequent number map.Remove(arr[i] - 1); } // If there is no last consequent of // arr[i] then put arr[i] else { if(!map.ContainsKey(arr[i])) map.Add(arr[i], 1); } } int max = int.MinValue; foreach(KeyValuePair<int, int> entry in map) { if(entry.Value > max) { max = entry.Value; } } return max;}// Driver codepublic static void Main(String []args){ // Take input from user int []arr = {3, 10, 3, 11, 4, 5, 6, 7, 8, 12}; int n = arr.Length; Console.WriteLine(longIncrConseqSubseq(arr, n));}}// This code is contributed by gauravrajput1 |
Javascript
<script>// JavaScript program to find length of the// longest increasing subsequence// whose adjacent element differ by 1// function that returns the length of the// longest increasing subsequence// whose adjacent element differ by 1function longestSubsequence(a, n){ // stores the index of elements var mp = new Map(); // stores the length of the longest // subsequence that ends with a[i] var dp = Array(n).fill(0); var maximum = -1000000000; // iterate for all element for (var i = 0; i < n; i++) { // if a[i]-1 is present before i-th index if (mp.has(a[i] - 1)) { // last index of a[i]-1 var lastIndex = mp.get(a[i] - 1) - 1; // relation dp[i] = 1 + dp[lastIndex]; } else dp[i] = 1; // stores the index as 1-index as we need to // check for occurrence, hence 0-th index // will not be possible to check mp.set(a[i], i + 1); // stores the longest length maximum = Math.max(maximum, dp[i]); } return maximum;}// Driver Codevar a = [3, 10, 3, 11, 4, 5, 6, 7, 8, 12];var n = a.length;document.write( longestSubsequence(a, n));</script> |
Output
6
Complexity Analysis:
- Time Complexity: O(N), as we are using a loop to traverse N times.
- Auxiliary Space: O(N), as we are using extra space for dp and map m.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
