Divide array in two maximum equal length arrays of similar and dissimilar elements
Given an array arr of natural numbers up to n, find the maximum size for which array arr can be divided into two equal-sized arrays, such that the first array contains all same elements while the second array contains all distinct elements.
Examples:
Input :n = 8, arr[] ={7, 3, 7, 1, 7, 7}
Output :
Maximum size is : 3
arr1[] ={7, 7, 7}
arr2[] ={1, 3, 7}
Explanation :
It is possible to construct two arrays of size 3.
The first array is [7, 7, 7] and the second array is [1, 3, 7].Input : n = 7, arr[] ={1, 2, 1, 5, 1, 6, 7, 2}
Output :
Maximum size is : 3
arr1[] ={1, 1, 1}
arr2[] ={2, 5, 6}
Approach:
To solve the problem mentioned above the main idea is to use hashing to find the frequency of every element present in the array.
- Find the element with maximum frequency present in the array arr[] using hash vector v.
- Find the total unique elements present in array arr[].
- There are two cases for the element with maximum frequency: the maximum frequency element will go to the first array then the sizes of the array are at most diff1 – 1 and max1 correspondingly. Otherwise, at least one element of maximum frequency goes to the second array and the sizes are at most diff1 and max1 ? 1 correspondingly. Then find the max-size to which array can be splitted as max(min(diff1 ? 1, max1), min(diff1, max1 ? 1)).
- Find the first array of similar elements using max_size and element with maximum frequency max1.
- Find the second array of unique elements using max_size and hash vector v.
Below is the implementation of the above approach:
C++
// C++ program to find the max-size to which // an array can be divided into 2 equal parts // such that one part contains unique elements // while another contains similar elements #include <bits/stdc++.h> using namespace std; // Function to find the max-size to which an array // can be divided into 2 equal parts void Solve(int arr[], int size, int n) { vector<int> v(n + 1); // Vector to find the frequency of // each element of array for (int i = 0; i < size; i++) v[arr[i]]++; // Find the maximum frequency // element present in array arr[] int max1 = (max_element(v.begin(), v.end()) - v.begin()); // Find total unique elements // present in array arr[] int diff1 = n + 1 - count(v.begin(), v.end(), 0); // Find the Max-Size to which // an array arr[] can be splitted int max_size = max(min(v[max1] - 1, diff1), min(v[max1], diff1 - 1)); cout << "Maximum size is :" << max_size << "\n"; // Find the first array // containing same elements cout << "The First Array Is : \n"; for (int i = 0; i < max_size; i++) { cout << max1 << " "; v[max1] -= 1; } cout << "\n"; // Find the second array // containing unique elements cout << "The Second Array Is : \n"; for (int i = 0; i < (n + 1); i++) { if (v[i] > 0) { cout << i << " "; max_size--; } if (max_size < 1) break; } cout << "\n"; } // Driver code int main() { // initialise n int n = 7; // array declaration int arr[] = { 1, 2, 1, 5, 1, 6, 7, 2 }; // size of array int size = sizeof(arr) / sizeof(arr[0]); Solve(arr, size, n); return 0; } |
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Java
// Java program to find the // max-size to which an array // can be divided into 2 equal parts // such that one part contains unique // elements while another contains // similar elements import java.io.*; import java.lang.*; import java.util.*; class GFG{ // Function to find the max-size to // which an array can be divided into // 2 equal parts static void Solve(int arr[], int size, int n) { int[] v = new int[n + 1]; // Array to find the frequency of // each element of array for (int i = 0; i < size; i++) v[arr[i]]++; // Find the index maximum frequency // element present in array arr[] int max1 = -1, mx = -1; for (int i = 0; i < v.length; i++) { if (v[i] > mx) { mx = v[i]; max1 = i; } } // Find total unique elements // present in array arr[] int cnt = 0; for (int i : v) { if (i == 0) ++cnt; } int diff1 = n + 1 - cnt; // Find the Max-Size to which // an array arr[] can be splitted int max_size = Math.max(Math.min(v[max1] - 1, diff1), Math.min(v[max1], diff1 - 1)); System.out.println("Maximum size is: " + max_size); // Find the first array // containing same elements System.out.println("First Array is"); for (int i = 0; i < max_size; i++) { System.out.print(max1 + " "); v[max1] -= 1; } System.out.println(); // Find the second array // containing unique elements System.out.println("The Second Array Is :"); for (int i = 0; i < (n + 1); i++) { if (v[i] > 0) { System.out.print(i + " "); max_size--; } if (max_size < 1) break; } System.out.println(); } // Driver Code public static void main(String[] args) { // initialise n int n = 7; // array declaration int arr[] = new int[] {1, 2, 1, 5, 1, 6, 7, 2}; // size of array int size = arr.length; Solve(arr, size, n); } } // This code is contributed by Sri_srajit |
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Python3
# Python3 program to find the max-size to which # an array can be divided into 2 equal parts # such that one part contains unique elements # while another contains similar elements # Function to find the max-size to which an # array can be divided into 2 equal parts def Solve(arr, size, n): v = [0] * (n + 1); # Vector to find the frequency of # each element of list for i in range(size): v[arr[i]] += 1 # Find the maximum frequency # element present in list arr max1 = max(set(arr), key = v.count) # Find total unique elements # present in list arr diff1 = n + 1 - v.count(0) # Find the Max-Size to which # an array arr[] can be splitted max_size = max(min(v[max1] - 1, diff1), min(v[max1], diff1 - 1)) print("Maximum size is :", max_size) # Find the first array # containing same elements print("The First Array Is : ") for i in range(max_size): print(max1, end = " ") v[max1] -= 1 print() # Find the second array # containing unique elements print("The Second Array Is : ") for i in range(n + 1): if (v[i] > 0): print(i, end = " ") max_size -= 1 if (max_size < 1): break print() # Driver code if __name__ == "__main__": # Initialise n n = 7 # Array declaration arr = [ 1, 2, 1, 5, 1, 6, 7, 2 ] # Size of array size = len(arr) Solve(arr, size, n) # This code is contributed by chitranayal |
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C#
// C# program to find the max-size // to which an array can be divided // into 2 equal parts such that one // part contains unique elements // while another contains similar // elements using System; class GFG{ // Function to find the max-size to // which an array can be divided into // 2 equal parts static void Solve(int []arr, int size, int n) { int[] v = new int[n + 1]; // Array to find the frequency of // each element of array for(int i = 0; i < size; i++) v[arr[i]]++; // Find the index maximum frequency // element present in array arr[] int max1 = -1, mx = -1; for(int i = 0; i < v.Length; i++) { if (v[i] > mx) { mx = v[i]; max1 = i; } } // Find total unique elements // present in array arr[] int cnt = 0; foreach(int i in v) { if (i == 0) ++cnt; } int diff1 = n + 1 - cnt; // Find the Max-Size to which // an array arr[] can be splitted int max_size = Math.Max(Math.Min(v[max1] - 1, diff1), Math.Min(v[max1], diff1 - 1)); Console.Write("Maximum size is :" + max_size + "\n"); // Find the first array // containing same elements Console.Write("The First Array Is :\n"); for(int i = 0; i < max_size; i++) { Console.Write(max1 + " "); v[max1] -= 1; } Console.Write("\n"); // Find the second array // containing unique elements Console.Write("The Second Array Is :\n"); for(int i = 0; i < (n + 1); i++) { if (v[i] > 0) { Console.Write(i + " "); max_size--; } if (max_size < 1) break; } Console.Write("\n"); } // Driver Code public static void Main(string[] args) { // Initialise n int n = 7; // Array declaration int []arr = new int[] { 1, 2, 1, 5, 1, 6, 7, 2 }; // Size of array int size = arr.Length; Solve(arr, size, n); } } // This code is contributed by rutvik_56 |
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Output:
Maximum size is :3 The First Array Is : 1 1 1 The Second Array Is : 2 5 6
Time Complexity :O(N)
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