给定一个整数数组，请在删除给定元素后找到最小的数字。如果重复元素，我们将为数组中包含要删除元素的每个实例删除一个实例(从原始数组中删除)。
例子：
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
方法 ：
- 将所有要从数组中删除的数字插入到哈希图中，以便我们可以检查数组元素是否在O(1)时也出现在Delete-array中。
- 将最小的最小值min初始化为INT_MAX。
- 遍历数组。检查元素是否存在于哈希图中。
- 如果存在，则将其从哈希映射中删除；否则，将其与min变量进行比较，如果元素的值小于min值，则更改其值。
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 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 29C#
// 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 mp = new Dictionary();
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 输出：
12
时间复杂度– O(N)