# Partition array into minimum number of equal length subsets consisting of a single distinct value

Given an array arr[] of size N, the task is to print the minimum count of equal length subsets the array can be partitioned into such that each subset contains only a single distinct element

Examples:

Input: arr[] = { 1, 2, 3, 4, 4, 3, 2, 1 }
Output:
Explanation:
Possible partition of the array is { {1, 1}, {2, 2}, {3, 3}, {4, 4} }.
Therefore, the required output is 4.

Input: arr[] = { 1, 1, 1, 2, 2, 2, 3, 3 }
Output:
Explanation:
Possible partition of the array is { {1}, {1}, {1}, {2}, {2}, {2}, {3}, {3} }.
Therefore, the required output is 8.

Naive Approach: The simplest approach to solve the problem is to store the frequency of each distinct array element, iterate over the range [N, 1] using the variable i, and check if the frequency of all distinct elements of the array is divisible by i or not. If found to be true, then print the value of (N / i)

Time Complexity: O(N2).
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach the idea is to use the concept of GCD. Follow the steps below to solve the problem:

• Initialize a map, say freq, to store the frequency of each distinct element of the array.
• Initialize a variable, say, FreqGCD, to store the GCD of frequency of each distinct element of the array.
• Traverse the map to find the value of FreqGCD.
• Finally, print the value of (N) % FreqGCD.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement` `// the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find the minimum count of subsets` `// by partitioning the array with given conditions` `int` `CntOfSubsetsByPartitioning(``int` `arr[], ``int` `N)` `{` `    ``// Store frequency of each` `    ``// distinct element of the array` `    ``unordered_map<``int``, ``int``> freq;`   `    ``// Traverse the array` `    ``for` `(``int` `i = 0; i < N; i++) {`   `        ``// Update frequency` `        ``// of arr[i]` `        ``freq[arr[i]]++;` `    ``}`   `    ``// Stores GCD of frequency of` `    ``// each distinct element of the array` `    ``int` `freqGCD = 0;` `    ``for` `(``auto` `i : freq) {`   `        ``// Update freqGCD` `        ``freqGCD = __gcd(freqGCD, i.second);` `    ``}`   `    ``return` `(N) / freqGCD;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `arr[] = { 1, 2, 3, 4, 4, 3, 2, 1 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << CntOfSubsetsByPartitioning(arr, N);` `    ``return` `0;` `}`

## Java

 `// Java program to implement` `// the above approach` `import` `java.util.*;`   `class` `GFG{`   `// Function to find the minimum count of subsets` `// by partitioning the array with given conditions` `static` `int` `CntOfSubsetsByPartitioning(``int` `arr[], ``int` `N)` `{` `    ``// Store frequency of each` `    ``// distinct element of the array` `    ``HashMap freq = ``new` `HashMap<>();`   `    ``// Traverse the array` `    ``for` `(``int` `i = ``0``; i < N; i++) {`   `        ``// Update frequency` `        ``// of arr[i]` `        ``if``(freq.containsKey(arr[i])){` `            ``freq.put(arr[i], freq.get(arr[i])+``1``);` `        ``}` `        ``else``{` `            ``freq.put(arr[i], ``1``);` `        ``}` `    ``}`   `    ``// Stores GCD of frequency of` `    ``// each distinct element of the array` `    ``int` `freqGCD = ``0``;` `    ``for` `(Map.Entry i : freq.entrySet()) {`   `        ``// Update freqGCD` `        ``freqGCD = __gcd(freqGCD, i.getValue());` `    ``}`   `    ``return` `(N) / freqGCD;` `}` `  `  `// Recursive function to return gcd of a and b  ` `static` `int` `__gcd(``int` `a, ``int` `b)  ` `{  ` ` ``return` `b == ``0``? a:__gcd(b, a % b);     ` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``4``, ``3``, ``2``, ``1` `};` `    ``int` `N = arr.length;` `    ``System.out.print(CntOfSubsetsByPartitioning(arr, N));` `}` `}`   `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 program to implement` `# the above approach` `from` `math ``import` `gcd`   `# Function to find the minimum count ` `# of subsets by partitioning the array ` `# with given conditions` `def` `CntOfSubsetsByPartitioning(arr, N):` `    `  `    ``# Store frequency of each` `    ``# distinct element of the array` `    ``freq ``=` `{}`   `    ``# Traverse the array` `    ``for` `i ``in` `range``(N):` `        `  `        ``# Update frequency` `        ``# of arr[i]` `        ``freq[arr[i]] ``=` `freq.get(arr[i], ``0``) ``+` `1`   `    ``# Stores GCD of frequency of` `    ``# each distinct element of the array` `    ``freqGCD ``=` `0` `    `  `    ``for` `i ``in` `freq:` `        `  `        ``# Update freqGCD` `        ``freqGCD ``=` `gcd(freqGCD, freq[i])`   `    ``return` `(N) ``/``/` `freqGCD`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``arr ``=` `[ ``1``, ``2``, ``3``, ``4``, ``4``, ``3``, ``2``, ``1` `]` `    ``N ``=` `len``(arr)` `    `  `    ``print``(CntOfSubsetsByPartitioning(arr, N))`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program to implement` `// the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{`   `// Function to find the minimum count of subsets` `// by partitioning the array with given conditions` `static` `int` `CntOfSubsetsByPartitioning(``int` `[]arr, ``int` `N)` `{` `    ``// Store frequency of each` `    ``// distinct element of the array` `    ``Dictionary<``int``,``int``> freq = ``new` `Dictionary<``int``,``int``>();`   `    ``// Traverse the array` `    ``for` `(``int` `i = 0; i < N; i++) {`   `        ``// Update frequency` `        ``// of arr[i]` `        ``if``(freq.ContainsKey(arr[i])){` `            ``freq[arr[i]] = freq[arr[i]]+1;` `        ``}` `        ``else``{` `            ``freq.Add(arr[i], 1);` `        ``}` `    ``}`   `    ``// Stores GCD of frequency of` `    ``// each distinct element of the array` `    ``int` `freqGCD = 0;` `    ``foreach` `(KeyValuePair<``int``,``int``> i ``in` `freq) {`   `        ``// Update freqGCD` `        ``freqGCD = __gcd(freqGCD, i.Value);` `    ``}`   `    ``return` `(N) / freqGCD;` `}` `  `  `// Recursive function to return gcd of a and b  ` `static` `int` `__gcd(``int` `a, ``int` `b)  ` `{  ` ` ``return` `b == 0? a:__gcd(b, a % b);     ` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``int` `[]arr = { 1, 2, 3, 4, 4, 3, 2, 1 };` `    ``int` `N = arr.Length;` `    ``Console.Write(CntOfSubsetsByPartitioning(arr, N));` `}` `}`   ` ``// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`4`

Time Complexity: O(N * log(M)), where M is the smallest element of the array
Auxiliary Space: O(N)

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next