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: 4
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: 8
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 <bits/stdc++.h>using namespace std;// Function to find the minimum count of subsets// by partitioning the array with given conditionsint 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 Codeint 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 approachimport java.util.*;class GFG{// Function to find the minimum count of subsets// by partitioning the array with given conditionsstatic int CntOfSubsetsByPartitioning(int arr[], int N){ // Store frequency of each // distinct element of the array HashMap<Integer,Integer> 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<Integer,Integer> 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 Codepublic 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 approachfrom math import gcd# Function to find the minimum count # of subsets by partitioning the array # with given conditionsdef 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 Codeif __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 approachusing System;using System.Collections.Generic;class GFG{// Function to find the minimum count of subsets// by partitioning the array with given conditionsstatic 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 Codepublic 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
<script>// JavaScript program to implement// the above approach// Function to find the minimum count of subsets// by partitioning the array with given conditionsfunction CntOfSubsetsByPartitioning(arr, N) { // Store frequency of each // distinct element of the array var freq = {}; // Traverse the array for(var i = 0; i < N; i++) { // Update frequency // of arr[i] if (freq.hasOwnProperty(arr[i])) { freq[arr[i]] = freq[arr[i]] + 1; } else { freq[arr[i]] = 1; } } // Stores GCD of frequency of // each distinct element of the array var freqGCD = 0; for(const [key, value] of Object.entries(freq)) { // Update freqGCD freqGCD = __gcd(freqGCD, value); } return parseInt(N / freqGCD);}// Recursive function to return gcd of a and bfunction __gcd(a, b){ return b == 0 ? a : __gcd(b, a % b);}// Driver Codevar arr = [ 1, 2, 3, 4, 4, 3, 2, 1 ];var N = arr.length;document.write(CntOfSubsetsByPartitioning(arr, N));// This code is contributed by rdtank</script> |
Output:
4
Time Complexity: O(N * log(M)), where M is the smallest element of the array
Auxiliary Space: O(N)
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