给定一个大小为N的数组arr[] ，任务是打印该数组可以划分成的等长子集的最小计数，使得每个子集只包含一个不同的元素
例子：
Input: arr[] = { 1, 2, 3, 4, 4, 3, 2, 1 }
Output: 4
Explanation:
Possible partition of the array is { {1, 1}, {2, 2}, {3, 3}, {4, 4} }.
Therefore, the required output is 4.
Input: arr[] = { 1, 1, 1, 2, 2, 2, 3, 3 }
Output: 8
Explanation:
Possible partition of the array is { {1}, {1}, {1}, {2}, {2}, {2}, {3}, {3} }.
Therefore, the required output is 8.
朴素方法：解决问题的最简单方法是存储每个不同数组元素的频率，使用变量i在范围[N, 1] 上迭代，并检查数组中所有不同元素的频率是否可被整除我与否。如果发现为真，则打印(N / i)的值。
时间复杂度： O(N 2 )。
辅助空间： O(N)
高效的方法：优化上述方法的想法是使用 GCD 的概念。请按照以下步骤解决问题：
- 初始化一个映射，比如freq ，以存储数组中每个不同元素的频率。
- 初始化一个变量，比如FreqGCD ，以存储数组中每个不同元素的频率的 GCD。
- 遍历地图以找到FreqGCD的值。
- 最后，打印(N) % FreqGCD的值。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to find the minimum count of subsets
// by partitioning the array with given conditions
int CntOfSubsetsByPartitioning(int arr[], int N)
{
// Store frequency of each
// distinct element of the array
unordered_map freq;
// Traverse the array
for (int i = 0; i < N; i++) {
// Update frequency
// of arr[i]
freq[arr[i]]++;
}
// Stores GCD of frequency of
// each distinct element of the array
int freqGCD = 0;
for (auto i : freq) {
// Update freqGCD
freqGCD = __gcd(freqGCD, i.second);
}
return (N) / freqGCD;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4, 4, 3, 2, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << CntOfSubsetsByPartitioning(arr, N);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to find the minimum count of subsets
// by partitioning the array with given conditions
static int CntOfSubsetsByPartitioning(int arr[], int N)
{
// Store frequency of each
// distinct element of the array
HashMap freq = new HashMap<>();
// Traverse the array
for (int i = 0; i < N; i++) {
// Update frequency
// of arr[i]
if(freq.containsKey(arr[i])){
freq.put(arr[i], freq.get(arr[i])+1);
}
else{
freq.put(arr[i], 1);
}
}
// Stores GCD of frequency of
// each distinct element of the array
int freqGCD = 0;
for (Map.Entry i : freq.entrySet()) {
// Update freqGCD
freqGCD = __gcd(freqGCD, i.getValue());
}
return (N) / freqGCD;
}
// Recursive function to return gcd of a and b
static int __gcd(int a, int b)
{
return b == 0? a:__gcd(b, a % b);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4, 4, 3, 2, 1 };
int N = arr.length;
System.out.print(CntOfSubsetsByPartitioning(arr, N));
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program to implement
# the above approach
from math import gcd
# Function to find the minimum count
# of subsets by partitioning the array
# with given conditions
def CntOfSubsetsByPartitioning(arr, N):
# Store frequency of each
# distinct element of the array
freq = {}
# Traverse the array
for i in range(N):
# Update frequency
# of arr[i]
freq[arr[i]] = freq.get(arr[i], 0) + 1
# Stores GCD of frequency of
# each distinct element of the array
freqGCD = 0
for i in freq:
# Update freqGCD
freqGCD = gcd(freqGCD, freq[i])
return (N) // freqGCD
# Driver Code
if __name__ == '__main__':
arr = [ 1, 2, 3, 4, 4, 3, 2, 1 ]
N = len(arr)
print(CntOfSubsetsByPartitioning(arr, N))
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the minimum count of subsets
// by partitioning the array with given conditions
static int CntOfSubsetsByPartitioning(int []arr, int N)
{
// Store frequency of each
// distinct element of the array
Dictionary freq = new Dictionary();
// Traverse the array
for (int i = 0; i < N; i++) {
// Update frequency
// of arr[i]
if(freq.ContainsKey(arr[i])){
freq[arr[i]] = freq[arr[i]]+1;
}
else{
freq.Add(arr[i], 1);
}
}
// Stores GCD of frequency of
// each distinct element of the array
int freqGCD = 0;
foreach (KeyValuePair i in freq) {
// Update freqGCD
freqGCD = __gcd(freqGCD, i.Value);
}
return (N) / freqGCD;
}
// Recursive function to return gcd of a and b
static int __gcd(int a, int b)
{
return b == 0? a:__gcd(b, a % b);
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 2, 3, 4, 4, 3, 2, 1 };
int N = arr.Length;
Console.Write(CntOfSubsetsByPartitioning(arr, N));
}
}
// This code is contributed by 29AjayKumar 输出：
4时间复杂度： O(N * log(M))，其中 M 是数组的最小元素
辅助空间： O(N)
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