从给定数组的任何排列中重复之间的最小距离最大化

给定一个数组arr [] ，该数组由范围[1，N]中的N个正整数组成，任务是从给定数组的任何排列中找到元素的任何连续重复之间的最大最小距离。

例子：

Input: arr[] = {1, 2, 1, 3} 
Output: 3 
Explanation: The maximum possible distance between the repetition is 3, from the permutation {1, 2, 3, 1} or {1, 3, 2, 1}.
Input: arr[] = {1, 2, 3, 4} 
Output: 0

方法：请按照以下步骤解决问题：

1. 存储每个数组元素的频率。
2. 找到包含最大频率的元素，例如maxFreqElement 。
3. 计算具有最大频率(例如maxFreqCount)的元素的出现次数。
4. 通过公式(N- maxFreqCount)/(maxFreqElement-1)计算所需距离

下面是上述方法的实现。

C++

// C++ Program to implement 
// the above approach 
#include  
using namespace std; 
int findMaxLen(vector& a) 
{ 
  
    // Size of the array 
    int n = a.size(); 
  
    // Stores the frequency of 
    // array elements 
    int freq[n + 1]; 
    memset(freq, 0, sizeof freq); 
  
    for (int i = 0; i < n; ++i) { 
        freq[a[i]]++; 
    } 
  
    int maxFreqElement = INT_MIN; 
    int maxFreqCount = 1; 
  
    for (int i = 1; i <= n; ++i) { 
  
        // Find the highest frequency 
        // in the array 
        if (freq[i] > maxFreqElement) { 
            maxFreqElement = freq[i]; 
            maxFreqCount = 1; 
        } 
  
        // Increase count of max frequent element 
        else if (freq[i] == maxFreqElement) 
            maxFreqCount++; 
    } 
  
    int ans; 
  
    // If no repetition is present 
    if (maxFreqElement == 1) 
        ans = 0; 
    else { 
        // Find the maximum distance 
        ans = ((n - maxFreqCount) 
            / (maxFreqElement - 1)); 
    } 
  
    // Return the max distance 
    return ans; 
} 
  
// Driver Code 
int main() 
{ 
  
    vector a = { 1, 2, 1, 2 }; 
    cout << findMaxLen(a) << endl; 
  
}

Java

// Java program to implement 
// the above approach 
class GFG{
      
static int findMaxLen(int a[], int n) 
{ 
      
    // Stores the frequency of 
    // array elements 
    int freq[] = new int[n + 1]; 
  
    for(int i = 0; i < n; ++i)
    { 
        freq[a[i]]++; 
    } 
  
    int maxFreqElement = Integer.MIN_VALUE; 
    int maxFreqCount = 1; 
  
    for(int i = 1; i <= n; ++i)
    { 
          
        // Find the highest frequency 
        // in the array 
        if (freq[i] > maxFreqElement)
        { 
            maxFreqElement = freq[i]; 
            maxFreqCount = 1; 
        } 
  
        // Increase count of max frequent element 
        else if (freq[i] == maxFreqElement) 
            maxFreqCount++; 
    } 
  
    int ans; 
  
    // If no repetition is present 
    if (maxFreqElement == 1) 
        ans = 0; 
    else 
    {
          
        // Find the maximum distance 
        ans = ((n - maxFreqCount) / 
               (maxFreqElement - 1)); 
    } 
  
    // Return the max distance 
    return ans; 
} 
  
// Driver Code 
public static void main(String [] args) 
{ 
    int a[] = { 1, 2, 1, 2 }; 
    int n = a.length;
      
    System.out.print(findMaxLen(a, n));
}
}
  
// This code is contributed by chitranayal

Python3

# Python3 program to implement
# the above approach
import sys
  
def findMaxLen(a):
  
    # Size of the array
    n = len(a)
  
    # Stores the frequency of
    # array elements
    freq = [0] * (n + 1)
  
    for i in range(n):
        freq[a[i]] += 1
  
    maxFreqElement = -sys.maxsize - 1
    maxFreqCount = 1
  
    for i in range(1, n + 1):
  
        # Find the highest frequency
        # in the array
        if(freq[i] > maxFreqElement):
            maxFreqElement = freq[i]
            maxFreqCount = 1
  
        # Increase count of max frequent element
        elif(freq[i] == maxFreqElement):
            maxFreqCount += 1
  
    # If no repetition is present
    if(maxFreqElement == 1):
        ans = 0
    else:
          
        # Find the maximum distance
        ans = ((n - maxFreqCount) // 
               (maxFreqElement - 1))
  
    # Return the max distance
    return ans
  
# Driver Code
a = [ 1, 2, 1, 2 ]
  
# Function call
print(findMaxLen(a))
  
# This code is contributed by Shivam Singh

C#

// C# program to implement
// the above approach
using System;
class GFG{
  
    static int findMaxLen(int[] a, int n)
    {
  
        // Stores the frequency of
        // array elements
        int[] freq = new int[n + 1];
  
        for (int i = 0; i < n; ++i) 
        {
            freq[a[i]]++;
        }
      
        int maxFreqElement = int.MinValue;
        int maxFreqCount = 1;
      
        for (int i = 1; i <= n; ++i) 
        {
  
            // Find the highest frequency
            // in the array
            if (freq[i] > maxFreqElement) 
            {
                maxFreqElement = freq[i];
                maxFreqCount = 1;
            }
  
            // Increase count of max 
            // frequent element
            else if (freq[i] == maxFreqElement)
                maxFreqCount++;
        }
  
        int ans;
  
        // If no repetition is present
        if (maxFreqElement == 1)
            ans = 0;
        else
        {
  
            // Find the maximum distance
            ans = ((n - maxFreqCount) / 
                   (maxFreqElement - 1));
        }
  
        // Return the max distance
        return ans;
    }
  
    // Driver Code
    public static void Main(String[] args)
    {
        int[] a = {1, 2, 1, 2};
        int n = a.Length;
        Console.Write(findMaxLen(a, n));
    }
}
  
// This code is contributed by Amit Katiyar

输出：

2

时间复杂度： O(N)
辅助空间： O(N)