制作加泰罗尼亚语序列所需的最小更改

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📜 制作加泰罗尼亚语序列所需的最小更改

📅 最后修改于: 2021-05-04 23:51:48 🧑 作者: Mango

给定N个整数元素的数组arr [] ，任务是更改此数组的最小元素数，以使其包含加泰罗尼亚语序列的前N个项。因此，找到所需的最小更改。

前几个加泰罗尼亚语数字是1、1、2、5、14、42、132、429、1430、4862等。

例子：

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

为什么编程需要懂一点英语

方法：

采取无序多重集。在此多重集中插入加泰罗尼亚语序列的前N个词。
从左到右遍历数组。检查数组元素是否在多集中。如果存在，则从多重集中删除该元素。
遍历数组后，所需的最小更改将等于多集的大小。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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 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::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 s = new HashSet();
  
        // 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 s = new HashSet();
  
        // 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

= 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
?>

输出：