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++

// 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;
}

Java

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

Python3

# 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

// 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

PHP

<?php
// PHP implementation of the approach 
$MAX = 1000;

// To store first N Catalan numbers 
$catalan = array_fill(0, $MAX, 0);

// Function to find first n Catalan numbers 
function catalanDP($n) 
{
    global $catalan;
    
    // Initialize first two values in table 
    $catalan[0] = $catalan[1] = 1;

    // Filong entries in catalan[] 
    // using recursive formula 
    for ($i = 2; $i <= $n; $i++) 
    {
        $catalan[$i] = 0;
        for ($j = 0; $j < $i; $j++) 
        {
            $catalan[$i] += $catalan[$j] * 
                            $catalan[$i - $j - 1];
        }
    }
}

// Function to return the minimum 
// changes required 
function CatalanSequence($arr, $n) 
{
    global $catalan;
    
    // Find first n Catalan Numbers 
    catalanDP($n);

    $s = array();

    // a and b are first two 
    // Catalan Sequence numbers 
    $a = $b = 1;

    // Insert first n catalan elements to set 
    array_push($s, $a);
    if ($n >= 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
?>

JavaScript

<script>

// Javascript implementation of the approach
var MAX = 100000

// To store first N Catalan numbers
var catalan = Array(MAX);

// Function to find first n Catalan numbers
function catalanDP(n)
{
    // Initialize first two values in table
    catalan[0] = catalan[1] = 1;

    // Fill entries in catalan[] using recursive formula
    for (var i = 2; i <= n; i++) {
        catalan[i] = 0;
        for (var j = 0; j < i; j++)
            catalan[i] += catalan[j] * catalan[i - j - 1];
    }
}

// Function to return the minimum changes required
function CatalanSequence(arr, n)
{
    // Find first n Catalan Numbers
    catalanDP(n);

    var s = [];

    // a and b are first two
    // Catalan Sequence numbers
    var a = 1, b = 1;
    var c;

    // push first n catalan elements to set
    s.push(a);
    if (n >= 2)
        s.push(b);

    for (var i = 2; i < n; i++) {
        s.push(catalan[i]);
    }

    s.sort((a,b)=>b-a);
    for(var i =0; i<n; i++)
    {
        // If catalan element is present
        // in the array then remove it from set
        if(s.includes(arr[i]))
        {
            s.pop(arr[i]);
        }
    }

    // Return the remaining number of
    // elements in the set
    return s.length;
}

// Driver code
var arr = [1, 1, 2, 5, 41 ];
var n = arr.length;
document.write( CatalanSequence(arr, n));

</script>

Output:

Minimum changes required to make a Catalan Sequence

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