Maximize minimum distance between repetitions from any permutation of the given Array

Given an array arr[], consisting of N positive integers in the range [1, N], the task is to find the largest minimum distance between any consecutive repetition of an element from any permutation of the given array.

Examples:

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

Approach: Follow the steps below to solve the problem:  

1. Store the frequency of each array element.
2. Find the element which contains the maximum frequency, say maxFreqElement.
3. Count the number of occurrences of elements having a maximum frequency, say maxFreqCount.
4. Calculate the required distance by the equation (N- maxFreqCount)/( maxFreqElement- 1))

Below is the implementation of the above approach.

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
int findMaxLen(vector<int>& 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<int> 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

Javascript

<script>

// Javascript Program to implement
// the above approach

function findMaxLen(a)
{

    // Size of the array
    var n = a.length;

    // Stores the frequency of
    // array elements
    var freq = Array(n+1).fill(0);

    var i;
    for(i = 0; i < n; ++i) {
        freq[a[i]]++;
    }

    var maxFreqElement = -2147483648;
    var maxFreqCount = 1;

    for (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++;
    }

    var 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
    var a = [1, 2, 1, 2];
    document.write(findMaxLen(a));

</script>

Output:

2

Time Complexity: O(N) 
Auxiliary Space: O(N)