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

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

Last Updated: 02-09-2020

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.

  • Python3

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// 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; 
} 
Output:  
2

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

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