Java Program for Maximize elements using another array

Last Updated : 24 Mar, 2023
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Given two arrays with size n, maximize the first array by using the elements from the second array such that the new array formed contains n greatest but unique elements of both the arrays giving the second array priority (All elements of second array appear before first array). The order of appearance of elements is kept same in output as in input.
Examples:

Input : arr1[] = {2, 4, 3} 
arr2[] = {5, 6, 1} 
Output : 5 6 4 
As 5, 6 and 4 are maximum elements from two arrays giving second array higher priority. Order of elements is same in output as in input.
Input : arr1[] = {7, 4, 8, 0, 1} 
arr2[] = {9, 7, 2, 3, 6} 
Output : 9 7 6 4 8

Approach : We create an auxiliary array of size 2*n and store the elements of 2nd array in auxiliary array, and then we will store elements of 1st array in it. After that we will sort auxiliary array in decreasing order. To keep the order of elements according to input arrays we will use hash table. We will store 1st n largest unique elements of auxiliary array in hash table. Now we traverse the second array and store that elements of second array in auxiliary array that are present in hash table. Similarly we will traverse first array and store the elements that are present in hash table. In this way we get n unique and largest elements from both the arrays in auxiliary array while keeping the order of appearance of elements same.
Below is the implementation of above approach :

Java 
// Java program to print the maximum elements
// giving second array higher priority
import java.util.*;

class GFG 
{

// Function to maximize array elements
static void maximizeArray(int[] arr1,int[] arr2)
{
    // auxiliary array arr3 to store 
    // elements of arr1 & arr2
    int arr3[] = new int[10]; 
    for(int i = 0; i < arr3.length; i++)
    {
        //arr2 has high priority
        arr3[i] = 0;
    }
    
    // Arraylist to store n largest
    // unique elements
    ArrayList<Integer> al = new ArrayList<Integer>();
    
    for(int i = 0; i < arr2.length; i++)
    {
        if(arr3[arr2[i]] == 0) 
        { 
            // to avoid repetition of digits of arr2 in arr3
            arr3[arr2[i]] = 2;
            
            // simultaneously setting arraylist to 
            // preserve order of arr2 and arr3
            al.add(arr2[i]); 
        }
    }
    
    for(int i = 0; i < arr1.length; i++)
    {
        if(arr3[arr1[i]] == 0)
        {
            // if digit is already present in arr2
            // then priority is arr2
            arr3[arr1[i]] = 1;
            
            // simultaneously setting arraylist to
            // preserve order of arr1
            al.add(arr1[i]); 
        }
    }

    // to get only highest n elements(arr2+arr1)
    // and remove others from arraylist
    int count = 0;
    for(int j = 9; j >= 0; j--)
    {
        if(count < arr1.length & 
          (arr3[j] == 2 || arr3[j] == 1))
        {
            // to not allow those elements 
            // which are absent in both arrays
            count++;
        }
        else
        {
            al.remove(Integer.valueOf(j));
        }
    }

    int i = 0;
    for(int x:al)
    {
        arr1[i++] = x;
    }
}

// Function to print array elements
static void printArray(int[] arr)
{
    for(int x:arr)
    {
        System.out.print(x + " ");
    }
}

// Driver Code
public static void main(String args[])
{
    int arr1[] = {7, 4, 8, 0, 1};
    int arr2[] = {9, 7, 2, 3, 6};
    maximizeArray(arr1,arr2);
    printArray(arr1);
}
}

// This code is contributed by KhwajaBilkhis

Output: 
9 7 6 4 8

Time complexity: O(n * log n).
Auxiliary Space: O(n) as list has been created.

Please refer complete article on Maximize elements using another array for more details!


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