Longest Monotonically Increasing Subsequence Size (N log N): Simple implementation

Given an array of random numbers, find the longest monotonically increasing subsequence (LIS) in the array.
If you want to understand the O(NlogN) approach, it’s explained very clearly here.
In this post, a simple and time-saving implementation of O(NlogN) approach using stl is discussed. Below is the code for LIS O(NlogN) :
 

C

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// A simple C++ implementation to find LIS
#include<iostream>
#include<algorithm>
#include<set>
using namespace std;
 
// Return length of LIS in arr[] of size N
int lis(int arr[], int N)
{
    int i;
    set<int> s;
    set<int>::iterator k;
    for (i=0; i<N; i++)
    {
        // Check if the element was actually inserted
        // An element in set is not inserted if it is
        // already present. Please see
        if (s.insert(arr[i]).second)
        {
            // Find the position of inserted element in iterator k
            k = s.find(arr[i]);
 
            k++;  // Find the next greater element in set
 
            // If the new element is not inserted at the end, then
            // remove the greater element next to it (This is tricky)
            if (k!=s.end()) s.erase(k);
        }
    }
 
    // Note that set s may not contain actual LIS, but its size gives
    // us the length of LIS
    return s.size();
}
 
int main()
{
    int arr[] = {8, 9, 12, 10, 11};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << lis(arr, n)<< endl;
}

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Java

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// Java implementation
// to find LIS
import java.util.*;
class GFG{
 
// Return length of LIS
// in arr[] of size N
static int lis(int arr[],
               int N)
{
  int i;
  HashSet<Integer> s = new HashSet<>();
  for (i = 0; i < N; i++)
  {
    // Check if the element
    // was actually inserted
    // An element in set is
    // not inserted if it is
    // already present. Please see
    // org/set-insert-function-in-c-stl/
    int k = 0;
    int size = s.size();
    if (s.add(arr[i]))
    {
      // Find the position of
      // inserted element in iterator k
      if(s.contains(arr[i]))
        k++; 
       
      // Find the next
      // greater element in set
      // If the new element is not
      // inserted at the end, then
      // remove the greater element
      // next to it.
      if (size == s.size())
        s.remove(k + 1);
    }
  }
 
  // Note that set s may not contain
  // actual LIS, but its size gives
  // us the length of LIS
  return s.size();
}
 
public static void main(String[] args)
{
  int arr[] = {8, 9, 12, 10, 11};
  int n = arr.length;
  System.out.print(lis(arr, n) + "\n");
}
}
 
// This code is contributed by gauravrajput1

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Output: 

4

This article is contributed by Raj Kumar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

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Improved By : warlockx, GauravRajput1

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