Printing longest Increasing consecutive subsequence

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

Examples:

Input : a[] = {3, 10, 3, 11, 4, 5, 6, 7, 8, 12} 
Output : 3 4 5 6 7 8 
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 : 6 7 8 9 10 
Explanation: 6, 7, 8, 9, 10 is the longest increasing subsequence 

We have discussed how to find length of Longest Increasing consecutive subsequence. To print the subsequence, we store index of last element. Then we print consecutive elements ending with last element.

Given below is the implementation 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
void 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
    int index = -1;
    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
        if (maximum < dp[i]) {
            maximum = dp[i];
            index = i;
        }
    }

    // We know last element of sequence is
    // a[index]. We also know that length
    // of subsequence is "maximum". So We
    // print these many consecutive elements
    // starting from "a[index] - maximum + 1"
    // to a[index].
    for (int curr = a[index] - maximum + 1;
         curr <= a[index]; curr++)
        cout << curr << " ";
}

// Driver Code
int main()
{
    int a[] = { 3, 10, 3, 11, 4, 5, 6, 7, 8, 12 };
    int n = sizeof(a) / sizeof(a[0]);
    longestSubsequence(a, n);
    return 0;
}

Java

// Java program to find length of the
// longest increasing subsequence
// whose adjacent element differ by
import java.util.HashMap;

class GFG
{

    // function that returns the length of the
    // longest increasing subsequence
    // whose adjacent element differ by 1
    public static void longestSubsequence(int[] a,
                                          int n)
    {

        // stores the index of elements
        HashMap<Integer,
                Integer> mp = new HashMap<>();

        // stores the length of the longest
        // subsequence that ends with a[i]
        int[] dp = new int[n];

        int maximum = Integer.MIN_VALUE;

        // iterate for all element
        int index = -1;
        for(int i = 0; i < n; i++)
        {

            // if a[i]-1 is present before i-th index
            if (mp.get(a[i] - 1) != null)
            {

                // last index of a[i]-1
                int 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.put(a[i], i +  1);

            // stores the longest length
            if (maximum < dp[i])
            {
                maximum = dp[i];
                index = i;
            }
        }

        // We know last element of sequence is
        // a[index]. We also know that length
        // of subsequence is "maximum". So We
        // print these many consecutive elements
        // starting from "a[index] - maximum + 1"
        // to a[index].
        for (int curr = a[index] - maximum + 1;
            curr <= a[index]; curr++)
            System.out.print(curr + " ");
    }

    // Driver Code
    public static void main(String[] args)
    {
        int[] a = { 3, 10, 3, 11, 4,
                    5, 6, 7, 8, 12 };
        int n = a.length;
        longestSubsequence(a, n);
    }
}

// This code is contributed by sanjeev2552

Python3

# Python 3 program to find length of
# the longest increasing subsequence
# whose adjacent element differ by 1
import sys

# function that returns the length
# of the longest increasing subsequence
# whose adjacent element differ by 1
def longestSubsequence(a, n):

    # stores the index of elements
    mp = {i:0 for i in range(13)}

    # stores the length of the longest
    # subsequence that ends with a[i]
    dp = [0 for i in range(n)]

    maximum = -sys.maxsize - 1

    # iterate for all element
    index = -1
    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
        if (maximum < dp[i]):
            maximum = dp[i]
            index = i

    # We know last element of sequence is
    # a[index]. We also know that length
    # of subsequence is "maximum". So We
    # print these many consecutive elements
    # starting from "a[index] - maximum + 1"
    # to a[index].
    for curr in range(a[index] - maximum + 1,
                      a[index] + 1, 1):
        print(curr, end = " ")

# Driver Code
if __name__ == '__main__':
    a = [3, 10, 3, 11, 4, 5,
                6, 7, 8, 12]
    n = len(a)
    longestSubsequence(a, n)

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find length of the
// longest increasing subsequence
// whose adjacent element differ by
using System;
using System.Collections.Generic;

class GFG
{

    // function that returns the length of the
    // longest increasing subsequence
    // whose adjacent element differ by 1
    static void longestSubsequence(int[] a, int n)
    {

        // stores the index of elements
        Dictionary<int,
                   int> mp = new Dictionary<int,
                                            int>();

        // stores the length of the longest
        // subsequence that ends with a[i]
        int[] dp = new int[n];

        int maximum = -100000000;

        // iterate for all element
        int index = -1;
        for(int i = 0; i < n; i++)
        {

            // if a[i]-1 is present before i-th index
            if (mp.ContainsKey(a[i] - 1) == true)
            {

                // 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
            if (maximum < dp[i])
            {
                maximum = dp[i];
                index = i;
            }
        }

        // We know last element of sequence is
        // a[index]. We also know that length
        // of subsequence is "maximum". So We
        // print these many consecutive elements
        // starting from "a[index] - maximum + 1"
        // to a[index].
        for (int curr = a[index] - maximum + 1;
            curr <= a[index]; curr++)
            Console.Write(curr + " ");
    }

    // Driver Code
    static void Main()
    {
        int[] a = { 3, 10, 3, 11, 4,
                    5, 6, 7, 8, 12 };
        int n = a.Length;
        longestSubsequence(a, n);
    }
}

// This code is contributed by mohit kumar

Javascript

<script>
// Javascript program to find length of the
// longest increasing subsequence
// whose adjacent element differ by

    // 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
        let mp = new Map();

        // stores the length of the longest
        // subsequence that ends with a[i]
        let dp = new Array(n);

        let maximum = Number.MIN_VALUE;

        // iterate for all element
        let index = -1;
        for(let i = 0; i < n; i++)
        {

            // if a[i]-1 is present before i-th index
            if (mp.get(a[i] - 1) != null)
            {

                // last index of a[i]-1
                let 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
            if (maximum < dp[i])
            {
                maximum = dp[i];
                index = i;
            }
        }

        // We know last element of sequence is
        // a[index]. We also know that length
        // of subsequence is "maximum". So We
        // print these many consecutive elements
        // starting from "a[index] - maximum + 1"
        // to a[index].
        for (let curr = a[index] - maximum + 1;
            curr <= a[index]; curr++)
            document.write(curr + " ");
    }

    // Driver Code
    let a=[3, 10, 3, 11, 4,
                    5, 6, 7, 8, 12 ];

    let n = a.length;
    longestSubsequence(a, n);

// This code is contributed by patel2127
</script>

Output

3 4 5 6 7 8 

Complexity Analysis:

• Time Complexity: O(n), as we are using a loop to traverse n times and in each traversal.
• Auxiliary Space: O(n), as we are using extra space for dp and mp.