删除给定元素后，查找k个最大数字
原文：https://www.geeksforgeeks.org/find-the-k-largest-numbers-after-deleting-the-given-elements/
给定一个整数数组，请在删除给定元素后找到k个最大数。 如果重复元素，请为包含要删除元素的数组中存在的元素的每个实例删除一个实例。
假定删除n个元素后将剩余至少k个元素。
示例：
输入：array[] = {5, 12, 33, 4, 56, 12, 20}, del[] = {12, 56, 5}, k = 3
输出：33 20 12
说明：删除后将保留{33, 4, 12, 20}。 从中打印出前 3 个最高的元素。
方法：
-
将所有要从数组中删除的数字插入到哈希映射中，以便我们可以检查数组元素是否在
O(1)时也出现在删除数组中。 -
遍历数组。 检查元素是否存在于哈希映射中。
-
如果存在，请从哈希映射中删除它。
-
否则，将其插入最大堆中。
-
插入所有要删除的元素（不包括要删除的元素）后，从最大堆中弹出
k个元素。
C++
#include "iostream"
#include "queue"
#include "unordered_map"
using namespace std;
// Find k maximum element from arr[0..m-1] after deleting
// elements from del[0..n-1]
void findElementsAfterDel(int arr[], int m, int del[],
int n, int k)
{
// 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
priority_queue<int> heap;
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
heap.push(arr[i]);
}
// Print top k elements in the heap
for (int i = 0; i < k; ++i) {
cout << heap.top() << " ";
// Pop the top element
heap.pop();
}
}
int main()
{
int array[] = { 5, 12, 33, 4, 56, 12, 20 };
int m = sizeof(array) / sizeof(array[0]);
int del[] = { 12, 56, 5 };
int n = sizeof(del) / sizeof(del[0]);
int k = 3;
findElementsAfterDel(array, m, del, n, k);
return 0;
}
Python3
# Python3 program to find the k maximum
# number from the array after n deletions
import math as mt
# Find k maximum element from arr[0..m-1]
# after deleting elements from del[0..n-1]
def findElementsAfterDel(arr, m, dell, n, k):
# 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
heap = list()
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 push it
else:
heap.append(arr[i])
heap.sort()
heap = heap[::-1]
return heap[:k]
# Driver code
array = [5, 12, 33, 4, 56, 12, 20]
m = len(array)
dell = [12, 4, 56, 5 ]
n = len(dell)
k = 3
print(*findElementsAfterDel(array, m, dell, n, k))
# This code is contributed by
# mohit kumar 29
输出：
33 20 12
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