给定一个项目数组,第 i 个索引元素表示项目 id,并给定一个数字 m,任务是删除 m 个元素,以便应留下最小的不同 id。打印不同 id 的数量。
例子:
Input : arr[] = { 2, 2, 1, 3, 3, 3}
m = 3
Output : 1
Remove 1 and both 2's.So, only 3 will be
left that's why distinct id is 1.
Input : arr[] = { 2, 4, 1, 5, 3, 5, 1, 3}
m = 2
Output : 3
Remove 2 and 4 completely. So, remaining ids
are 1, 3 and 5 i.e. 3
询问:摩根士丹利
1- 计算元素的出现次数并将它们存储在哈希中。
2- 对哈希进行排序。
3- 开始从哈希中删除元素。
4- 返回散列中剩余的值数。
C++
// C++ program for above implementation
#include
using namespace std;
// Function to find distintc id's
int distinctIds(int arr[], int n, int mi)
{
unordered_map m;
vector > v;
int count = 0;
// Store the occurrence of ids
for (int i = 0; i < n; i++)
m[arr[i]]++;
// Store into the vector second as first and vice-versa
for (auto it = m.begin(); it != m.end(); it++)
v.push_back(make_pair(it->second, it->first));
// Sort the vector
sort(v.begin(), v.end());
int size = v.size();
// Start removing elements from the beginning
for (int i = 0; i < size; i++) {
// Remove if current value is less than
// or equal to mi
if (v[i].first <= mi) {
mi -= v[i].first;
count++;
}
// Return the remaining size
else
return size - count;
}
return size - count;
}
// Driver code
int main()
{
int arr[] = { 2, 3, 1, 2, 3, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
int m = 3;
cout << distinctIds(arr, n, m);
return 0;
}
Java
//Java program for Minimum number of
//distinct elements after removing m items
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
public class DistinctIds
{
// Function to find distintc id's
static int distinctIds(int arr[], int n, int mi)
{
Map m = new HashMap();
int count = 0;
int size = 0;
// Store the occurrence of ids
for (int i = 0; i < n; i++)
{
// If the key is not add it to map
if (m.containsKey(arr[i]) == false)
{
m.put(arr[i], 1);
size++;
}
// If it is present then increase the value by 1
else m.put(arr[i], m.get(arr[i]) + 1);
}
// Start removing elements from the beginning
for (Entry mp:m.entrySet())
{
// Remove if current value is less than
// or equal to mi
if (mp.getKey() <= mi)
{
mi -= mp.getKey();
count++;
}
// Return the remaining size
else return size - count;
}
return size - count;
}
//Driver method to test above function
public static void main(String[] args)
{
// TODO Auto-generated method stub
int arr[] = {2, 3, 1, 2, 3, 3};
int m = 3;
System.out.println(distinctIds(arr, arr.length, m));
}
}
//This code is contributed by Sumit Ghosh
Python3
# Python program for above implementation
# Function to find distintc id's
def distinctIds(arr, n, mi):
m = {}
v = []
count = 0
# Store the occurrence of ids
for i in range(n):
if arr[i] in m:
m[arr[i]] += 1
else:
m[arr[i]] = 1
# Store into the list value as key and vice-versa
for i in m:
v.append([m[i],i])
v.sort()
size = len(v)
# Start removing elements from the beginning
for i in range(size):
# Remove if current value is less than
# or equal to mi
if (v[i][0] <= mi):
mi -= v[i][0]
count += 1
else: # Return the remaining size
return size - count
return size - count
# Driver code
arr = [ 2, 3, 1, 2, 3, 3 ]
n = len(arr)
m = 3
# To display the result
print(distinctIds(arr, n, m))
# This code is contributed by rohitsingh07052
C#
// C# program for Minimum number of
// distinct elements after removing m items
using System;
using System.Collections.Generic;
class GFG
{
// Function to find distintc id's
static int distinctIds(int[] arr, int n, int mi)
{
Dictionary m = new Dictionary();
int count = 0;
int size = 0;
// Store the occurrence of ids
for (int i = 0; i < n; i++)
{
// If the key is not add it to map
if (m.ContainsKey(arr[i]) == false)
{
m[arr[i]] = 1;
size++;
}
// If it is present then increase the value by 1
else
{
m[arr[i]]++;
}
}
// Start removing elements from the beginning
foreach(KeyValuePair mp in m)
{
// Remove if current value is less than
// or equal to mi
if (mp.Key <= mi)
{
mi -= mp.Key;
count++;
}
}
return size - count;
}
// Driver code
static void Main()
{
// TODO Auto-generated method stub
int[] arr = {2, 3, 1, 2, 3, 3};
int m = 3;
Console.WriteLine(distinctIds(arr, arr.Length, m));
}
}
// This code is contributed by divyeshrabadiya07
Javascript
输出:
1
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