Longest Increasing consecutive subsequence

Given N elements, write a program that prints the length of the longest increasing subsequence whose adjacent element difference is one. 

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 any particular element, find the length of the subsequence starting from the first element. Print the longest length of the subsequence thus formed. The time complexity of this approach will be O(n). 

C++

#include <bits/stdc++.h>
using namespace std;
int LongestSubsequence(int a[], int n)
{
    // initialize x to the first element of the array;
    int x = a[0];
    // initialize count to Zero
    int count = 0, y = 0;
    // iterate for all elements
    for (int i = 0; i < n; i++) {
        if (a[i] == (x + y)) {
            count++;
            y++;
        }
    }
    return count;
}
int main()
{
    int a[] = { 6, 7, 8, 3, 4, 5, 9, 10 };
    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)
  {

    // initialize x to the first element of the array;
    int x = a[0];

    // initialize count to Zero
    int count = 0, y = 0;

    // iterate for all elements
    for (int i = 0; i < n; i++) {
      if (a[i] == (x + y)) {
        count++;
        y++;
      }
    }
    return count;
  }
  public static void main(String[] args) {
    int[] a = {6, 7, 8, 3, 4, 5, 9, 10};
    int n = a.length;
    System.out.println(LongestSubsequence(a, n));
  }
}

// This code contributed by Ajax

Python3

def LongestSubsequence(a, n):
    # initialize x to the first element of the array;
    x = a[0]

    # initialize count to Zero
    count = 0
    y = 0

    # iterate for all elements
    for i in range(n):
        if a[i] == (x + y):
            count += 1
            y += 1
    return count

a = [6, 7, 8, 3, 4, 5, 9, 10]
n = len(a)
print(LongestSubsequence(a, n))

# This code is contributed by laxmisinde5t82.

C#

using System;

public class GFG {

    public static int LongestSubsequence(int[] a, int n)
    {

        // initialize x to the first element of the array;
        int x = a[0];

        // initialize count to Zero
        int count = 0, y = 0;

        // iterate for all elements
        for (int i = 0; i < n; i++) {
            if (a[i] == (x + y)) {
                count++;
                y++;
            }
        }
        return count;
    }
    public static void Main(string[] args)
    {
        int[] a = { 6, 7, 8, 3, 4, 5, 9, 10 };
        int n = a.Length;
        Console.WriteLine(LongestSubsequence(a, n));
    }
}

// This code contributed by Karandeep1234

Javascript

function LongestSubsequence(a, n)
{

    // initialize x to the first element of the array;
    var x = a[0];

    // initialize count to Zero
    var count = 0, y = 0;

    // iterate for all elements
    for (var i = 0; i < n; i++) {
        if (a[i] == (x + y)) {
            count++;
            y++;
        }
    }
    return count;
}

var a = [6, 7, 8, 3, 4, 5, 9, 10];
var n = a.length;
console.log(LongestSubsequence(a, n));

// This code contributed by Prasad Kandekar(prasad264)

Output

5

Time Complexity: O(n)
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 1
int 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 Code
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

// Java program to find length of the
// longest increasing subsequence
// whose adjacent element differ by 1

import 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 defaultdict
import 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 1
using 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 code
public 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 1
function 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 Code
var 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.