所需的最小段数，以便每个段具有不同的元素

给定一个整数数组，任务是找到数组元素可以分成的最小段数，使得所有段都包含不同的元素。

例子：

Input: n=6 ; Array: 1, 7, 4, 3, 3, 8
Output: 2
Explanation:
Optimal way to create segments here is {1, 7, 4, 3} {3, 8}
Clearly, the answer is the maximum frequency of any element within the array i.e. '2'.
as '3' is the element which appears the most in the array (twice).

Input : n=5 ; Array: 2, 2, 3, 3, 3, 5
Output : 3

方法：

• 最佳方法是将所有不同的元素放在一个段中。
• 然后，将所有其他出现多次的元素一一放在新段中，以便没有段包含重复的元素。
• 因此，答案是给定数组中任何元素的最大频率。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function that counts the
// minimum segments required
void CountSegments(int N, int a[])
{
    // all values are '0' initially
    int frequency[10001] = { 0 };
 
    // count of segments
    int c = 0;
 
    // store frequency of every element
    for (int i = 0; i < N; i++) {
        frequency[a[i]]++;
    }
 
    // find maximum frequency
    for (int i = 0; i <= 10000; i++)
        c = max(c, frequency[i]);
 
    cout << c << "\n";
}
 
// Driver code
int main()
{
    int N = 6;
    int a[] = { 1, 3, 4, 3, 2, 3 };
 
    CountSegments(N, a);
 
    return 0;
}

Java

// Java implementation of the approach
 
import java.io.*;
 
class GFG {
 
// Function that counts the
// minimum segments required
static void CountSegments(int N, int a[])
{
    // all values are '0' initially
    int frequency[] =  new int[10001];
 
    // count of segments
    int c = 0;
 
    // store frequency of every element
    for (int i = 0; i < N; i++) {
        frequency[a[i]]++;
    }
 
    // find maximum frequency
    for (int i = 0; i <= 10000; i++)
        c = Math.max(c, frequency[i]);
 
    System.out.println( c);
}
 
       // Driver code
    public static void main (String[] args) {
        int N = 6;
    int []a = { 1, 3, 4, 3, 2, 3 };
 
    CountSegments(N, a);
    }
}
 
// This Code is contributed by inder_verma..

Python 3

# Python3 implementation of the approach
 
# Function that counts the
# minimum segments required
def CountSegments(N, a):
     
    # all values are '0' initially
    frequency = [0] * 10001
 
    # count of segments
    c = 0
 
    # store frequency of every element
    for i in range(N) :
        frequency[a[i]] += 1
 
    # find maximum frequency
    for i in range(10001):
        c = max(c, frequency[i])
 
    print(c)
 
# Driver code
if __name__ == "__main__":
    N = 6
    a = [ 1, 3, 4, 3, 2, 3 ]
    CountSegments(N, a)
 
# This code is contributed
# by ChitraNayal

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function that counts the
// minimum segments required
static void CountSegments(int N, int []a)
{
    // all values are '0' initially
    int []frequency = new int[10001];
 
    // count of segments
    int c = 0;
 
    // store frequency of every element
    for (int i = 0; i < N; i++)
    {
        frequency[a[i]]++;
    }
 
    // find maximum frequency
    for (int i = 0; i <= 10000; i++)
        c = Math.Max(c, frequency[i]);
 
    Console.WriteLine( c);
}
 
// Driver code
public static void Main ()
{
    int N = 6;
    int []a = { 1, 3, 4, 3, 2, 3 };
 
    CountSegments(N, a);
}
}
 
// This code is contributed
// by inder_verma

Javascript

 
// js implementation of the approach
 
// Function that counts the
// minimum segments required
function CountSegments(N,a)
{
    // all values are '0' initially
    var frequency = new Array(10001).fill(0);
 
    // count of segments
    let c = 0;
 
    // store frequency of every element
    for (var i = 0; i < N; i++)
    {
        frequency[a[i]]++;
    }
 
    // find maximum frequency
    for (var i = 0; i <= 10000; i++)
        c = Math.max(c, frequency[i]);
 
    document.write(c);
}
 
// Driver code
 
var N = 6;
var a = [1, 3, 4, 3, 2, 3 ];
 
CountSegments(N, a);
 
 

输出：

3

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