Find the largest after deleting the given elements

Given an array of integers, find the largest number after deleting the given elements. In case of repeating elements, delete one instance for every instance of the element present in the array containing the elements to be deleted.

Examples:

Input : array[] = { 5, 12, 33, 4, 56, 12, 20 } 
del[] = { 12, 33, 56, 5 } 
Output : 20 
Explanation : We get {12, 20} after deleting given elements. Largest among remaining element is 20

Approach :

• Insert all the numbers in the hash map which are to be deleted from the array, so that we can check if the element in the array is also present in the Delete-array in O(1) time.
• Initialize largest number max to be INT_MIN.
• Traverse through the array. Check if the element is present in the hash map.
• If present, erase it from the hash map, else if not present compare it with max variable and change its value if the value of the element is greater than the max value.

Implementation:

C++

// 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

// 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

# 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#

// 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

Javascript

<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>

Output

20

Complexity Analysis:

• Time Complexity: O(max(m, n)
• Auxiliary Space: O(m + n)