Find the smallest after deleting given elements

Given an array of integers, find the smallest number after deleting given elements. In case of repeating elements, we delete one instance (from the original array) for every instance present in the array containing elements to be deleted.

Examples:

Input : Array = {5, 12, 33, 4, 56, 12, 20} 
To Delete = {12, 4, 56, 5} 
Output : 12 
After deleting given elements, array becomes {33, 12, 20} and minimum element becomes 12. Note that there are two occurrences of 12 and we delete one of them. 

Input : Array = {1, 20, 3, 4, 10} 
To Delete = {1, 4, 10} 
Output : 3

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 smallest number min to be INT_MAX. 
• 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 the min variable and change its value if the value of the element is less than the min value. 

Implementation:

C++

// C++ program to find the smallest number
// from the array after  n deletions
#include "climits"
#include "iostream"
#include "unordered_map"
using namespace std;

// Returns minimum element from arr[0..m-1] after deleting
// elements from del[0..n-1]
int findSmallestAfterDel(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 smallestElement
    int smallestElement = INT_MAX;

    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 smallestElement
        else
            smallestElement = min(smallestElement, arr[i]);
    }

    return smallestElement;
}

int main()
{
    int array[] = { 5, 12, 33, 4, 56, 12, 20 };
    int m = sizeof(array) / sizeof(array[0]);

    int del[] = { 12, 4, 56, 5 };
    int n = sizeof(del) / sizeof(del[0]);

    cout << findSmallestAfterDel(array, m, del, n);
    return 0;
}

Java

// Java program to find the smallest number
// from the array after n deletions
import java.util.*;

class GFG
{

// Returns minimum element from arr[0..m-1]
// after deleting elements from del[0..n-1]
static int findSmallestAfterDel(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 smallestElement
    int smallestElement = Integer.MAX_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 smallestElement
        else
            smallestElement = Math.min(smallestElement,
                                               arr[i]);
    }

    return smallestElement;
}

// Driver Code
public static void main(String[] args)
{
    int array[] = { 5, 12, 33, 4, 56, 12, 20 };
    int m = array.length;

    int del[] = { 12, 4, 56, 5 };
    int n = del.length;

    System.out.println(findSmallestAfterDel(array, m,
                                            del, n));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to find the smallest
# 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 findSmallestAfterDel(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 SmallestElement
    SmallestElement = 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 SmallestElement
        else:
            SmallestElement = min(SmallestElement,
                                           arr[i])

    return SmallestElement

# Driver code
array = [5, 12, 33, 4, 56, 12, 20]
m = len(array)

dell = [12, 4, 56, 5]
n = len(dell)

print(findSmallestAfterDel(array, m, dell, n))

# This code is contributed
# by mohit kumar 29

C#

// C# program to find the smallest number
// from the array after n deletions
using System;
using System.Collections.Generic;

class GFG
{

// Returns minimum element from arr[0..m-1]
// after deleting elements from del[0..n-1]
static int findSmallestAfterDel(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[del[i]] = mp[del[i]] + 1;
        }
        else
        {
            mp.Add(del[i], 1);
        }
    }

    // Initializing the smallestElement
    int smallestElement = int.MaxValue;
    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 smallestElement
        else
            smallestElement = Math.Min(smallestElement,
                                               arr[i]);
    }
    return smallestElement;
}

// Driver Code
public static void Main(String[] args)
{
    int []array = { 5, 12, 33, 4, 56, 12, 20 };
    int m = array.Length;

    int []del = { 12, 4, 56, 5 };
    int n = del.Length;

    Console.WriteLine(findSmallestAfterDel(array, m,
                                           del, n));
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// JavaScript program to find the smallest number
// from the array after n deletions

// Returns minimum element from arr[0..m-1]
// after deleting elements from del[0..n-1]
function findSmallestAfterDel(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 smallestElement
    let smallestElement = Number.MAX_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 smallestElement
        else
            smallestElement = Math.min(smallestElement,
                                               arr[i]);
    }

    return smallestElement;
}

// Driver Code
let array=[5, 12, 33, 4, 56, 12, 20];
let m = array.length;

let del=[12, 4, 56, 5];
let n = del.length;
document.write(findSmallestAfterDel(array, m,
                                            del, n));


// This code is contributed by patel2127

</script>

Output

12

Time Complexity: O(N)
Auxiliary Space: O(N)