Minimum changes required to make a Catalan Sequence

Given an array arr[] of N integer elements, the task is to change the minimum number of elements of this array such that it contains first N terms of the Catalan Sequence. Thus, find the minimum changes required.

First few Catalan numbers are 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, …..

Examples:

Input: arr[] = {4, 1, 2, 33, 213, 5}
Output: 3
We have to replace 4, 33, 213 with 1, 14, 42 to make first 6 terms of Catalan sequence.

Input: arr[] = {1, 1, 2, 5, 41}
Output: 1
Simply change 41 with 14

Approach:

  • Take an unordered multiset. Insert first N terms of Catalan sequence in this multiset.
  • Traverse the array from left to right. Check if the array element if present in the multiset. If it is present, then remove that element from the multiset.
  • After traversing the array, the minimum changes required will be equal to the size of the multiset.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define MAX 100000
#define ll long long int
  
// To store first N Catalan numbers
ll catalan[MAX];
  
// Function to find first n Catalan numbers
void catalanDP(ll n)
{
  
    // Initialize first two values in table
    catalan[0] = catalan[1] = 1;
  
    // Fill entries in catalan[] using recursive formula
    for (int i = 2; i <= n; i++) {
        catalan[i] = 0;
        for (int j = 0; j < i; j++)
            catalan[i] += catalan[j] * catalan[i - j - 1];
    }
}
  
// Function to return the minimum changes required
int CatalanSequence(int arr[], int n)
{
  
    // Find first n Catalan Numbers
    catalanDP(n);
  
    unordered_multiset<int> s;
  
    // a and b are first two
    // Catalan Sequence numbers
    int a = 1, b = 1;
    int c;
  
    // Insert first n catalan elements to set
    s.insert(a);
    if (n >= 2)
        s.insert(b);
  
    for (int i = 2; i < n; i++) {
        s.insert(catalan[i]);
    }
  
    unordered_multiset<int>::iterator it;
  
    for (int i = 0; i < n; i++) {
  
        // If catalan element is present
        // in the array then remove it from set
        it = s.find(arr[i]);
        if (it != s.end())
            s.erase(it);
    }
  
    // Return the remaining number of
    // elements in the set
    return s.size();
}
  
// Driver code
int main()
{
    int arr[] = { 1, 1, 2, 5, 41 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << CatalanSequence(arr, n);
  
    return 0;
}

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Java

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import java.util.HashSet;
  
// Java implementation of the approach 
class GFG1 
{
  
    static int MAX = 100000;
  
    // To store first N Catalan numbers 
    static long catalan[] = new long[MAX];
  
    // Function to find first n Catalan numbers 
    static void catalanDP(long n) 
    {
  
        // Initialize first two values in table 
        catalan[0] = catalan[1] = 1;
  
        // Filong entries in catalan[] 
        // using recursive formula 
        for (int i = 2; i <= n; i++) 
        {
            catalan[i] = 0;
            for (int j = 0; j < i; j++) 
            {
                catalan[i] += catalan[j] * catalan[i - j - 1];
            }
        }
    }
  
    // Function to return the minimum changes required 
    static int CatalanSequence(int arr[], int n) 
    {
  
        // Find first n Catalan Numbers 
        catalanDP(n);
  
        HashSet<Integer> s = new HashSet<Integer>();
  
        // a and b are first two 
        // Catalan Sequence numbers 
        int a = 1, b = 1;
        int c;
  
        // Insert first n catalan elements to set 
        s.add(a);
        if (n >= 2
        {
            s.add(b);
        }
  
        for (int i = 2; i < n; i++) 
        {
            s.add((int) catalan[i]);
        }
  
        for (int i = 0; i < n; i++) 
        {
  
            // If catalan element is present 
            // in the array then remove it from set 
            if (s.contains(arr[i])) 
            {
                s.remove(arr[i]);
            }
        }
  
        // Return the remaining number of 
        // elements in the set 
        return s.size();
    }
  
    // Driver code 
    public static void main(String[] args) 
    {
        int arr[] = {1, 1, 2, 5, 41};
        int n = arr.length;
  
        System.out.print(CatalanSequence(arr, n));
    }
  
// This code contributed by Rajput-Ji

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Python3

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# Python3 implementation of
# the approach 
MAX = 100000
  
# To store first N Catalan numbers 
catalan = [0] * MAX
  
# Function to find first n 
# Catalan numbers 
def catalanDP(n) : 
  
    # Initialize first two values 
    # in table 
    catalan[0] = catalan[1] = 1
  
    # Fill entries in catalan[] 
    # using recursive formula 
    for i in range(2, n + 1) :
        catalan[i] = 0
        for j in range(i) :
            catalan[i] += (catalan[j] * 
                           catalan[i - j - 1]); 
  
# Function to return the minimum 
# changes required 
def CatalanSequence(arr, n) :
      
    # Find first n Catalan Numbers 
    catalanDP(n); 
  
    s = set(); 
  
    # a and b are first two 
    # Catalan Sequence numbers 
    a = 1 ; b = 1
  
    # Insert first n catalan 
    # elements to set 
    s.add(a); 
    if (n >= 2) :
        s.add(b); 
  
    for i in range(2, n) :
        s.add(catalan[i]); 
      
    temp = set()
    for i in range(n) :
  
        # If catalan element is present 
        # in the array then remove it 
        # from set 
        if arr[i] in s :
            temp.add(arr[i])
      
    s = s - temp ;
      
    # Return the remaining number 
    # of elements in the set 
    return len(s); 
  
# Driver code 
if __name__ == "__main__" :
  
    arr = [1, 1, 2, 5, 41]; 
    n = len(arr)
  
    print(CatalanSequence(arr, n)); 
  
# This code is contributed by Ryuga

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C#

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// C# implementation of the approach 
using System;
using System.Collections.Generic;
  
class GFG1 
{
  
    static int MAX = 100000;
  
    // To store first N Catalan numbers 
    static long[] catalan = new long[MAX];
  
    // Function to find first n Catalan numbers 
    static void catalanDP(long n) 
    {
  
        // Initialize first two values in table 
        catalan[0] = catalan[1] = 1;
  
        // Filong entries in catalan[] 
        // using recursive formula 
        for (int i = 2; i <= n; i++) 
        {
            catalan[i] = 0;
            for (int j = 0; j < i; j++) 
            {
                catalan[i] += catalan[j] * catalan[i - j - 1];
            }
        }
    }
  
    // Function to return the minimum changes required 
    static int CatalanSequence(int []arr, int n) 
    {
  
        // Find first n Catalan Numbers 
        catalanDP(n);
  
        HashSet<int> s = new HashSet<int>();
  
        // a and b are first two 
        // Catalan Sequence numbers 
        int a = 1, b = 1;
  
        // Insert first n catalan elements to set 
        s.Add(a);
        if (n >= 2) 
        {
            s.Add(b);
        }
  
        for (int i = 2; i < n; i++) 
        {
            s.Add((int)catalan[i]);
        }
  
        for (int i = 0; i < n; i++) 
        {
  
            // If catalan element is present 
            // in the array then remove it from set 
            if (s.Contains(arr[i])) 
            {
                s.Remove(arr[i]);
            }
        }
  
        // Return the remaining number of 
        // elements in the set 
        return s.Count;
    }
  
    // Driver code 
    public static void Main() 
    {
        int []arr = {1, 1, 2, 5, 41};
        int n = arr.Length;
  
        Console.WriteLine(CatalanSequence(arr, n));
    }
  
// This code contributed by mits

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PHP

= 2)
{
array_push($s, $b);
}

for ($i = 2; $i < $n; $i++) { array_push($s, $catalan[$i]); } $s = array_unique($s); for ($i = 0; $i < $n; $i++) { // If catalan element is present // in the array then remove it from set if (in_array($arr[$i], $s)) { unset($s[array_search($arr[$i], $s)]); } } // Return the remaining number of // elements in the set return count($s); } // Driver code $arr = array(1, 1, 2, 5, 41); $n = count($arr); print(CatalanSequence($arr, $n)); // This code contributed by mits ?>

Output:

1


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