# Find the smallest after deleting given elements

• Difficulty Level : Easy
• Last Updated : 07 Sep, 2022

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 mp = ``new` `HashMap();``    ``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

 ``

Output

`12`

Time Complexity: O(N)

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