Longest increasing sub-sequence formed by concatenating array to itself N times

Given an array arr[] of size N, the task is to find the length of the longest increasing subsequence in the array formed by the concatenation of the arr[] to itself at the end N times.
Examples: 
 

Input: arr[] = {3, 2, 1}, N = 3 
Output:
Explanation: 
The array formed by the concatenation – 
{3, 2, 1, 3, 2, 1, 3, 2, 1} 
The longest increasing subsequence that can be formed from this array is of length 3 which is {1, 2, 3}

Input: N = 3 arr[] = {3, 1, 4} 
Output: 
Explanation: 
The array formed by concatenation – 
{3, 1, 4, 3, 1, 4, 3, 1, 4} 
The longest increasing subsequence that can be formed from this array is of length 3 which is {1, 3, 4} 

Naive Approach: 
The basic approach to solve this problem is to create the final array by concatenating the given array to itself N times, and then finding the longest increasing subsequence in it. 
Time Complexity: O(N2
Auxiliary Space: O(N2)

Efficient Approach: 
According to the efficient approach, any element that is present in the longest increasing subsequence can be present only once. It means that the repetition of the elements N times won’t affect the subsequence, but, any element can be chosen anytime. Therefore, it would be efficient to find the longest increasing subset in the array of length N, which can be found by finding all the unique elements of the array. 
Below is the algorithm for the efficient approach: 
Algorithm: 



  • Store the unique elements of the array in a map with (element, count) as the (key, value) pair.
  • For each element in the array 
    • If the current element is not present in the map, then insert it in the map, with count 1.
    • Otherwise, increment the count of the array elements in the array.
  • Find the length of the map which will be the desired answer.

For Example: 

Given Array be – {4, 4, 1}
Creating the map of unique elements: {(4, 2), (1, 1)} 
Length of the Map = 2 
Hence the required longest subsequence = 2

Below is the implementation of the above approach:
 

C++

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// C++ implementation to find the
// longest increasing subsequence
// in repeating element of array
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the LCS
int findLCS(int arr[], int n){
    unordered_map<int, int> mp;
     
    // Loop to create frequency array
    for (int i = 0; i < n; i++) {
        mp[arr[i]]++;
    }
    return mp.size();
}
 
// Driver code
int main()
{
    int n = 3;
    int arr[] = {3, 2, 1};
    cout<<findLCS(arr, n);
    return 0;
}

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Java

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// Java implementation to find the
// longest increasing subsequence
// in repeating element of array
import java.util.*;
 
class GFG{
 
// Function to find the LCS
static int findLCS(int arr[], int n)
{
    HashMap<Integer,
            Integer> mp = new HashMap<Integer,
                                      Integer>();
     
    // Loop to create frequency array
    for(int i = 0; i < n; i++)
    {
       if(mp.containsKey(arr[i]))
       {
           mp.put(arr[i], mp.get(arr[i]) + 1);
       }
       else
       {
           mp.put(arr[i], 1);
       }
    }
    return mp.size();
}
 
// Driver code
public static void main(String[] args)
{
    int n = 3;
    int arr[] = { 3, 2, 1 };
     
    System.out.print(findLCS(arr, n));
}
}
 
// This code is contributed by amal kumar choubey

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Python3

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# Python3 implementation to find the
# longest increasing subsequence
# in repeating element of array
 
# Function to find the LCS
def findLCS(arr, n):
     
    mp = {}
 
    # Loop to create frequency array
    for i in range(n):
        if arr[i] in mp:
            mp[arr[i]] += 1
        else:
            mp[arr[i]] = 1
             
    return len(mp)
 
# Driver code
n = 3
arr = [ 3, 2, 1 ]
 
print(findLCS(arr, n))
 
# This code is contributed by ng24_7

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C#

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// C# implementation to find the
// longest increasing subsequence
// in repeating element of array
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the LCS
static int findLCS(int []arr, int n)
{
    Dictionary<int,
               int> mp = new Dictionary<int,
                                        int>();
     
    // Loop to create frequency array
    for(int i = 0; i < n; i++)
    {
       if(mp.ContainsKey(arr[i]))
       {
           mp[arr[i]] = mp[arr[i]] + 1;
       }
       else
       {
           mp.Add(arr[i], 1);
       }
    }
    return mp.Count;
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 3;
    int []arr = { 3, 2, 1 };
     
    Console.Write(findLCS(arr, n));
}
}
 
// This code is contributed by amal kumar choubey

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Performance Analysis: 

  • Time Complexity: As in the above approach, there only one loop which takes O(N) time in worst case, Hence the Time Complexity will be O(N).
  • Space Complexity: As in the above approach, there one Hash map used which can take O(N) space in worst case, Hence the space complexity will be O(N)

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