# 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; ` `} ` |

*chevron_right*

*filter_none*

## 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
[tabbyending]
**Output: **

3 4 5 6 7 8

**Time Complexity:** O(n)

**Auxiliary Space:** O(n)

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