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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;
}

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)



This article is attributed to GeeksforGeeks.org

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