Encuentre el más grande después de eliminar los elementos dados

Dada una array de enteros, encuentre el número más grande después de eliminar los elementos dados. En caso de elementos repetidos, elimine una instancia por cada instancia del elemento presente en la array que contiene los elementos que se eliminarán.
Ejemplos:
 

Entrada: array [] = { 5, 12, 33, 4, 56, 12, 20 } 
del [] = { 12, 33, 56, 5 } 
Salida: 20 
Explicación: Obtenemos {12, 20} después de eliminar elementos dados . El más grande entre los elementos restantes es 20

Acercarse : 
 

• Inserte todos los números en el mapa hash que se eliminarán de la array, de modo que podamos verificar si el elemento en la array también está presente en la array Eliminar en tiempo O (1).
• Inicialice el número máximo máximo para que sea INT_MIN.
• Atraviesa la array. Compruebe si el elemento está presente en el mapa hash.
• Si está presente, bórrelo del mapa hash; de lo contrario, si no está presente, compárelo con la variable máxima y cambie su valor si el valor del elemento es mayor que el valor máximo.

// C++ program to find the largest number
// from the array after  n deletions
#include "climits"
#include "iostream"
#include "unordered_map"
using namespace std;
 
// Returns maximum element from arr[0..m-1] after deleting
// elements from del[0..n-1]
int findlargestAfterDel(int arr[], int m, int del[], int n)
{
    // Hash Map of the numbers to be deleted
    unordered_map<int, int> mp;
    for (int i = 0; i < n; ++i) {
 
        // Increment the count of del[i]
        mp[del[i]]++;
    }
 
    // Initializing the largestElement
    int largestElement = INT_MIN;
 
    for (int i = 0; i < m; ++i) {
 
        // Search if the element is present
        if (mp.find(arr[i]) != mp.end()) {
 
            // Decrement its frequency
            mp[arr[i]]--;
 
            // If the frequency becomes 0,
            // erase it from the map
            if (mp[arr[i]] == 0)
                mp.erase(arr[i]);
        }
 
        // Else compare it largestElement
        else
            largestElement = max(largestElement, arr[i]);
    }
 
    return largestElement;
}
 
int main()
{
    int array[] = { 5, 12, 33, 4, 56, 12, 20 };
    int m = sizeof(array) / sizeof(array[0]);
 
    int del[] = { 12, 33, 56, 5 };
    int n = sizeof(del) / sizeof(del[0]);
 
    cout << findlargestAfterDel(array, m, del, n);
    return 0;
}

// Java program to find the largest number
// from the array after n deletions
import java.util.*;
 
class GFG
{
 
// Returns maximum element from arr[0..m-1] after deleting
// elements from del[0..n-1]
static int findlargestAfterDel(int arr[], int m,
                               int del[], int n)
{
    // Hash Map of the numbers to be deleted
    HashMap<Integer,
            Integer> mp = new HashMap<Integer,
                                      Integer>();
    for (int i = 0; i < n; ++i)
    {
 
        // Increment the count of del[i]
        if(mp.containsKey(del[i]))
        {
            mp.put(del[i], mp.get(del[i]) + 1);
        }
        else
        {
            mp.put(del[i], 1);
        }
    }
 
    // Initializing the largestElement
    int largestElement = Integer.MIN_VALUE;
 
    for (int i = 0; i < m; i++)
    {
 
        // Search if the element is present
        if (mp.containsKey(arr[i]))
        {
 
            // Decrement its frequency
            mp.put(arr[i], mp.get(arr[i]) - 1);
 
            // If the frequency becomes 0,
            // erase it from the map
            if (mp.get(arr[i]) == 0)
                mp.remove(arr[i]);
        }
 
        // Else compare it largestElement
        else
            largestElement = Math.max(largestElement, arr[i]);
    }
    return largestElement;
}
 
// Driver Code
public static void main(String[] args)
{
    int array[] = { 5, 12, 33, 4, 56, 12, 20 };
    int m = array.length;
 
    int del[] = { 12, 33, 56, 5 };
    int n = del.length;
 
    System.out.println(findlargestAfterDel(array, m, del, n));   
}
}
 
// This code is contributed by Rajput-Ji

# Python3 program to find the largest
# number from the array after n deletions
import math as mt
 
# Returns maximum element from arr[0..m-1]
# after deleting elements from del[0..n-1]
def findlargestAfterDel(arr, m, dell, n):
 
    # Hash Map of the numbers
    # to be deleted
    mp = dict()
    for i in range(n):
         
        # Increment the count of del[i]
        if dell[i] in mp.keys():
            mp[dell[i]] += 1
        else:
            mp[dell[i]] = 1
             
    # Initializing the largestElement
    largestElement = -10**9
 
    for i in range(m):
         
        # Search if the element is present
        if (arr[i] in mp.keys()):
             
            # Decrement its frequency
            mp[arr[i]] -= 1
 
            # If the frequency becomes 0,
            # erase it from the map
            if (mp[arr[i]] == 0):
                mp.pop(arr[i])
                 
        # Else compare it largestElement
        else:
            largestElement = max(largestElement,
                                         arr[i])
 
    return largestElement
 
# Driver code
array = [5, 12, 33, 4, 56, 12, 20]
m = len(array)
 
dell = [12, 33, 56, 5]
n = len(dell)
 
print(findlargestAfterDel(array, m, dell, n))
 
# This code is contributed
# by mohit kumar 29

// C# program to find the largest number
// from the array after n deletions
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Returns maximum element from arr[0..m-1]
// after deleting elements from del[0..n-1]
static int findlargestAfterDel(int []arr, int m,
                               int []del, int n)
{
    // Hash Map of the numbers to be deleted
    Dictionary<int,
               int> mp = new Dictionary<int,
                                        int>();
    for (int i = 0; i < n; ++i)
    {
 
        // Increment the count of del[i]
        if(mp.ContainsKey(del[i]))
        {
            mp[arr[i]] = mp[arr[i]] + 1;
        }
        else
        {
            mp.Add(del[i], 1);
        }
    }
 
    // Initializing the largestElement
    int largestElement = int.MinValue;
 
    for (int i = 0; i < m; i++)
    {
 
        // Search if the element is present
        if (mp.ContainsKey(arr[i]))
        {
 
            // Decrement its frequency
            mp[arr[i]] = mp[arr[i]] - 1;
 
            // If the frequency becomes 0,
            // erase it from the map
            if (mp[arr[i]] == 0)
                mp.Remove(arr[i]);
        }
 
        // Else compare it largestElement
        else
            largestElement = Math.Max(largestElement,
                                             arr[i]);
    }
    return largestElement;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []array = { 5, 12, 33, 4, 56, 12, 20 };
    int m = array.Length;
 
    int []del = { 12, 33, 56, 5 };
    int n = del.Length;
 
    Console.WriteLine(findlargestAfterDel(array, m, del, n));
}
}
 
// This code is contributed by Princi Singh

<script>
// Javascript program to find the largest number
// from the array after n deletions
 
// Returns maximum element from arr[0..m-1] after deleting
// elements from del[0..n-1]
function findlargestAfterDel(arr,m,del,n)
{
    // Hash Map of the numbers to be deleted
    let mp = new Map();
    for (let i = 0; i < n; ++i)
    {
   
        // Increment the count of del[i]
        if(mp.has(del[i]))
        {
            mp.set(del[i], mp.get(del[i]) + 1);
        }
        else
        {
            mp.set(del[i], 1);
        }
    }
   
    // Initializing the largestElement
    let largestElement = Number.MIN_VALUE;
   
    for (let i = 0; i < m; i++)
    {
   
        // Search if the element is present
        if (mp.has(arr[i]))
        {
   
            // Decrement its frequency
            mp.set(arr[i], mp.get(arr[i]) - 1);
   
            // If the frequency becomes 0,
            // erase it from the map
            if (mp.get(arr[i]) == 0)
                mp.delete(arr[i]);
        }
   
        // Else compare it largestElement
        else
            largestElement = Math.max(largestElement, arr[i]);
    }
    return largestElement;
}
 
// Driver Code
let array=[5, 12, 33, 4, 56, 12, 20];
let m = array.length;
let del = [ 12, 33, 56, 5 ];
let n = del.length;
document.write(findlargestAfterDel(array, m, del, n));   
 
// This code is contributed by patel2127
</script>

Producción : 
 

20

Complejidad del tiempo – O(max(m, n)

Espacio Auxiliar – O(m + n)