给定由N个不同的整数和整数K组成的数组arr [] ，任务是找到可能的子集的最大大小，以使子集中的任何元素都不是该子集的任何其他元素的K倍(即，没有这样的对{ n，m}应该出现在子集中，使得m = n * K或n = m * K )。
例子：
Input: arr[] = {2, 8, 6, 5, 3}, K = 2
Output: 4
Explanation:
Only possible pair existing in the array with an element being K( = 2) times the other is {6, 3}.
Hence, all possible subsets which does not contain both the elements of the pair {6, 3} together can be considered.
Therefore, the longest possible subset can be of length 4.
Input: arr[] = {1, 4, 3, 2}, K = 3
output: 3
方法：
请按照以下步骤解决问题：
- 找到可能的对数，以使一个元素是给定数组中另一个元素的K倍
- 排序以递增要素的顺序数组。
- 遍历数组并将数组元素的频率存储在Map中。
- 初始化访问的数组以标记每个索引，无论该元素是否包含在子集中( 0 )( 1 )。
- 再次遍历数组，对于vis [i] = 0的每个索引，检查Map中是否存在arr [i] *K 。如果找到，则增加对的数量并设置vis [mp [arr [i] * K]] = 1 。
- 最后，打印N –对数作为答案。
下面是上述方法的实现：
C++
// C++ implementation of
// the above aproach
#include
#define ll long long
using namespace std;
// Function to find the maximum
// size of the required subset
int findMaxLen(vector& a, ll k)
{
// Size of the array
int n = a.size();
// Sort the array
sort(a.begin(), a.end());
// Stores which index is
// included or excluded
vector vis(n, 0);
// Stores the indices of
// array elements
map mp;
for (int i = 0; i < n; i++) {
mp[a[i]] = i;
}
// Count of pairs
int c = 0;
// Iterate through all
// the element
for (int i = 0; i < n; ++i) {
// If element is included
if (vis[i] == false) {
int check = a[i] * k;
// Check if a[i] * k is present
// in the array or not
if (mp.find(check) != mp.end()) {
// Increase count of pair
c++;
// Exclude the pair
vis[mp[check]] = true;
}
}
}
return n - c;
}
// Driver code
int main()
{
int K = 3;
vector arr = { 1, 4, 3, 2 };
cout << findMaxLen(arr, K);
} Java
// Java implementation of
// the above aproach
import java.util.*;
class GFG{
// Function to find the maximum
// size of the required subset
static int findMaxLen(int[] a, int k)
{
// Size of the array
int n = a.length;
// Sort the array
Arrays.sort(a);
// Stores which index is
// included or excluded
boolean []vis = new boolean[n];
// Stores the indices of
// array elements
HashMap mp = new HashMap();
for(int i = 0; i < n; i++)
{
mp.put(a[i], i);
}
// Count of pairs
int c = 0;
// Iterate through all
// the element
for(int i = 0; i < n; ++i)
{
// If element is included
if (vis[i] == false)
{
int check = a[i] * k;
// Check if a[i] * k is present
// in the array or not
if (mp.containsKey(check))
{
// Increase count of pair
c++;
// Exclude the pair
vis[mp.get(check)] = true;
}
}
}
return n - c;
}
// Driver code
public static void main(String[] args)
{
int K = 3;
int []arr = { 1, 4, 3, 2 };
System.out.print(findMaxLen(arr, K));
}
}
// This code is contributed by amal kumar choubey Python3
# Python3 implementation of
# the above approach
# Function to find the maximum
# size of the required subset
def findMaxLen(a, k):
# Size of the array
n = len(a)
# Sort the array
a.sort()
# Stores which index is
# included or excluded
vis = [0] * n
# Stores the indices of
# array elements
mp = {}
for i in range(n):
mp[a[i]] = i
# Count of pairs
c = 0
# Iterate through all
# the element
for i in range(n):
# If element is included
if(vis[i] == False):
check = a[i] * k
# Check if a[i] * k is present
# in the array or not
if(check in mp.keys()):
# Increase count of pair
c += 1
# Exclude the pair
vis[mp[check]] = True
return n - c
# Driver Code
if __name__ == '__main__':
K = 3
arr = [ 1, 4, 3, 2 ]
print(findMaxLen(arr, K))
# This code is contributed by Shivam SinghC#
// C# implementation of
// the above aproach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the maximum
// size of the required subset
static int findMaxLen(int[] a, int k)
{
// Size of the array
int n = a.Length;
// Sort the array
Array.Sort(a);
// Stores which index is
// included or excluded
bool []vis = new bool[n];
// Stores the indices of
// array elements
Dictionary mp = new Dictionary();
for(int i = 0; i < n; i++)
{
mp.Add(a[i], i);
}
// Count of pairs
int c = 0;
// Iterate through all
// the element
for(int i = 0; i < n; ++i)
{
// If element is included
if (vis[i] == false)
{
int check = a[i] * k;
// Check if a[i] * k is present
// in the array or not
if (mp.ContainsKey(check))
{
// Increase count of pair
c++;
// Exclude the pair
vis[mp[check]] = true;
}
}
}
return n - c;
}
// Driver code
public static void Main(String[] args)
{
int K = 3;
int []arr = { 1, 4, 3, 2 };
Console.Write(findMaxLen(arr, K));
}
}
// This code is contributed by gauravrajput1 输出：
3
时间复杂度： O(N)
辅助空间： O(N)