# 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:
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++ Program to implement  ` `// the above approach  ` `#include   ` `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 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 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# 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`

Output:
```2

```

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :