Split an array into equal length subsequences consisting of equal elements only

Last Updated : 19 Jan, 2021

Given an array arr[] of size N, the task is to check if it is possible to split the array arr[] into different subsequences of equal size such that each element of the subsequence are equal. If found to be true, then print “YES”. Otherwise, print “NO”.

Examples:

Input: arr[] = {1, 2, 3, 4, 4, 3, 2, 1}
Output: YES
Explanation: Possible partition: {1, 1}, {2, 2}, {3, 3}, {4, 4}.

Input: arr[] = {1, 1, 1, 2, 2, 2, 3, 3}
Output: NO

Approach: The idea is based on the following observation: Let the frequency of arr[i] be C_i, then these elements must be broken down into subsequences of X such that C_i % X = 0. This must be YES for every index i. To satisfy this, the value of X should be equal to the greatest common divisor(GCD) of all C_i(1≤i≤N). If X is greater than 1, then print YES otherwise print NO.

Follow the steps below to solve the problem:

Create a hashmap, mp, to store the frequencies of all the elements of the array, arr[].
Store the greatest common divisor of all the frequencies in mp in a variable X.
If X is greater than 1, then the answer is YES.
Otherwise, the answer is NO.

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 GCD
// of two numbers a and b
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function to check if it is possible to
// split the array into equal length subsequences
// such that all elements in the subsequence are equal
void splitArray(int arr[], int N)
{
 
    // Store frequencies of
    // array elements
    map<int, int> mp;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Update frequency of arr[i]
        mp[arr[i]]++;
    }
 
    // Store the GCD of frequencies
    // of all array elements
    int G = 0;
 
    // Traverse the map
    for (auto i : mp) {
 
        // Update GCD
        G = gcd(G, i.second);
    }
 
    // If the GCD is greater than 1,
    // print YES otherwise print NO
    if (G > 1)
        cout << "YES";
    else
        cout << "NO";
}
 
// Driver Code
int main()
{
 
    // Given array
    int arr[] = { 1, 2, 3, 4, 4, 3, 2, 1 };
 
    // Store the size of the array
    int n = sizeof(arr) / sizeof(arr[0]);
 
    splitArray(arr, n);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG 
{
 
    // Function to find the GCD
    // of two numbers a and b
    int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
 
    // Function to check if it is possible to
    // split the array into equal length subsequences
    // such that all elements in the subsequence are equal
    void splitArray(int arr[], int N)
    {
 
        // Store frequencies of
        // array elements
        TreeMap<Integer, Integer> mp
            = new TreeMap<Integer, Integer>();
 
        // Traverse the array
        for (int i = 0; i < N; i++)
        {
 
            // Update frequency of arr[i]
            if (mp.containsKey(arr[i])) 
            {
                mp.put(arr[i], mp.get(arr[i]) + 1);
            }
            else
            {
                mp.put(arr[i], 1);
            }
        }
 
        // Store the GCD of frequencies
        // of all array elements
        int G = 0;
 
        // Traverse the map
        for (Map.Entry<Integer, Integer> m :
             mp.entrySet())
        {
           
            // update gcd
            Integer i = m.getValue();
            G = gcd(G, i.intValue());
        }
 
        // If the GCD is greater than 1,
        // print YES otherwise print NO
        if (G > 1)
            System.out.print("YES");
        else
            System.out.print("NO");
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        // Given array
        int[] arr = new int[] { 1, 2, 3, 4, 4, 3, 2, 1 };
 
        // Store the size of the array
        int n = arr.length;
        new GFG().splitArray(arr, n);
    }
}
 
// This code is contributed by abhishekgiri1

Output:

YES

Time Complexity: O(N * log(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.

My Personal Notes

Related Articles

C++

Java