Related Articles

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

• Last Updated : 09 Jun, 2021

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);``    ``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)

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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up