Printing longest Increasing consecutive subsequence

08 Jun, 2022
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++
Java
Python3
C#
Javascript
// 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 1void 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 Codeint 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 program to find length of the// longest increasing subsequence// whose adjacent element differ byimport 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
# Python 3 program to find length of# the longest increasing subsequence# whose adjacent element differ by 1import sys # function that returns the length# of the longest increasing subsequence# whose adjacent element differ by 1def 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 Codeif __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# program to find length of the// longest increasing subsequence// whose adjacent element differ byusing 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
<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 
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. 
SURENDRA_GANGWAR
sanjeev2552
mohit kumar 29
patel2127
akshatagrawalmat18
rohitsingh07052
cpp-unordered_map
Arrays
Dynamic Programming
Hash
Arrays
Hash
Dynamic Programming
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