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 Ci, then these elements must be broken down into subsequences of X such that Ci % 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 Ci (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 for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the GCD// of two numbers a and bint 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 equalvoid 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 Codeint 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 for the above approachimport 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 |
Python3
# Python3 program for the above approachfrom collections import defaultdict# Function to find the GCD# of two numbers a and bdef gcd(a, 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 equaldef splitArray(arr, N): # Store frequencies of # array elements mp = defaultdict(int) # Traverse the array for i in range(N): # Update frequency of arr[i] mp[arr[i]] += 1 # Store the GCD of frequencies # of all array elements G = 0 # Traverse the map for i in mp: # Update GCD G = gcd(G, mp[i]) # If the GCD is greater than 1, # print YES otherwise print NO if (G > 1): print("YES") else: print("NO")# Driver Codeif __name__ == "__main__": # Given array arr = [ 1, 2, 3, 4, 4, 3, 2, 1 ] # Store the size of the array n = len(arr) splitArray(arr, n)# This code is contributed by chitranayal |
C#
// C# program for the above approachusing System;using System.Collections.Generic; using System.Linq;class GFG{ // Function to find the GCD// of two numbers a and bstatic 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 equalstatic void splitArray(int[] arr, int n){ // Store frequencies of // array elements Dictionary<int, int> mp = new Dictionary<int, int>(); // Traverse the array for(int i = 0; i < n; ++i) { // Update frequency of // each array element if (mp.ContainsKey(arr[i]) == true) mp[arr[i]] += 1; else mp[arr[i]] = 1; } // Store the GCD of frequencies // of all array elements int G = 0; // Traverse the map foreach (KeyValuePair<int, int> i in mp) { // Update GCD G = gcd(G, i.Value); } // If the GCD is greater than 1, // print YES otherwise print NO if (G > 1) Console.Write("YES"); else Console.Write("NO");}// Driver Codepublic static void Main(){ // Given array int[] arr = { 1, 2, 3, 4, 4, 3, 2, 1 }; // Store the size of the array int n = arr.Length; splitArray(arr, n);}}// This code is contributed by sanjoy_62. |
Javascript
<script>// Javascript program for the above approach// Function to find the GCD// of two numbers a and bfunction gcd(a, 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 equalfunction splitArray(arr, N){ // Store frequencies of // array elements var mp = new Map(); // Traverse the array for(var i = 0; i < N; i++) { // Update frequency of arr[i] if (mp.has(arr[i])) { mp.set(arr[i], mp.get(arr[i]) + 1); } else { mp.set(arr[i], 1); } } // Store the GCD of frequencies // of all array elements var G = 0; // Traverse the map mp.forEach((value, key) => { // Update GCD G = gcd(G, value); }); // If the GCD is greater than 1, // print YES otherwise print NO if (G > 1) document.write("YES"); else document.write("NO");}// Driver Code// Given arrayvar arr = [ 1, 2, 3, 4, 4, 3, 2, 1 ];// Store the size of the arrayvar n = arr.length;splitArray(arr, n);// This code is contributed by rutvik_56</script> |
Output:
YES
Time Complexity: O(N * log(N))
Auxiliary Space: O(N)
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