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Minimum changes required to make a Catalan Sequence
  • Last Updated : 12 May, 2021

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:
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:
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#




// 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: 
1

 




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