Maximize elements using another array

• Difficulty Level : Hard
• Last Updated : 25 May, 2021

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 :

C++

 // C++ program to print the maximum elements// giving second array higher priority#include using namespace std; // Compare function used to sort array// in decreasing orderbool compare(int a, int b){    return a > b;} // Function to maximize array elementsvoid maximizeArray(int arr1[], int arr2[],                                   int n){    // auxiliary array arr3 to store    // elements of arr1 & arr2    int arr3[2*n], k = 0;    for (int i = 0; i < n; i++)        arr3[k++] = arr1[i];    for (int i = 0; i < n; i++)        arr3[k++] = arr2[i];     // hash table to store n largest    // unique elements    unordered_set hash;     // sorting arr3 in decreasing order    sort(arr3, arr3 + 2 * n, compare);     // finding n largest unique elements    // from arr3 and storing in hash    int i = 0;    while (hash.size() != n) {         // if arr3 element not present in hash,        // then store this element in hash        if (hash.find(arr3[i]) == hash.end())            hash.insert(arr3[i]);                 i++;    }     // store that elements of arr2 in arr3    // that are present in hash    k = 0;    for (int i = 0; i < n; i++) {         // if arr2 element is present in hash,        // store it in arr3        if (hash.find(arr2[i]) != hash.end()) {            arr3[k++] = arr2[i];            hash.erase(arr2[i]);        }    }     // store that elements of arr1 in arr3    // that are present in hash    for (int i = 0; i < n; i++) {         // if arr1 element is present in hash,        // store it in arr3        if (hash.find(arr1[i]) != hash.end()) {            arr3[k++] = arr1[i];            hash.erase(arr1[i]);        }    }     // copying 1st n elements of arr3 to arr1    for (int i = 0; i < n; i++)        arr1[i] = arr3[i];   } // Function to print array elementsvoid printArray(int arr[], int n){    for (int i = 0; i < n; i++)        cout << arr[i] << " ";       cout << endl;} // Driver Codeint main(){    int array1[] = { 7, 4, 8, 0, 1 };    int array2[] = { 9, 7, 2, 3, 6 };    int size = sizeof(array1) / sizeof(array1);    maximizeArray(array1, array2, size);    printArray(array1, size);}

Java

 // Java program to print the maximum elements// giving second array higher priorityimport java.util.*; class GFG{ // Function to maximize array elementsstatic void maximizeArray(int[] arr1,int[] arr2){    // auxiliary array arr3 to store    // elements of arr1 & arr2    int arr3[] = new int;    for(int i = 0; i < arr3.length; i++)    {        //arr2 has high priority        arr3[i] = 0;    }         // Arraylist to store n largest    // unique elements    ArrayList al = new ArrayList();         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 elementsstatic void printArray(int[] arr){    for(int x:arr)    {        System.out.print(x + " ");    }} // Driver Codepublic 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

Python3

 # Python3 program to print the maximum elements# giving second array higher priority # Function to maximize array elementsdef maximizeArray(arr1, arr2, n):         # Auxiliary array arr3 to store    # elements of arr1 & arr2    arr3 =  * (2 * n)    k = 0         for i in range(n):        arr3[k] = arr1[i]        k += 1             for i in range(n):        arr3[k] = arr2[i]        k += 1     # Hash table to store n largest    # unique elements    hash = {}     # Sorting arr3 in decreasing order    arr3 = sorted(arr3)    arr3 = arr3[::-1]     # Finding n largest unique elements    # from arr3 and storing in hash    i = 0    while (len(hash) != n):         # If arr3 element not present in hash,        # then store this element in hash        if (arr3[i] not in hash):            hash[arr3[i]] = 1         i += 1     # Store that elements of arr2 in arr3    # that are present in hash    k = 0    for i in range(n):         # If arr2 element is present in        # hash, store it in arr3        if (arr2[i] in hash):            arr3[k] = arr2[i]            k += 1                         del hash[arr2[i]]     # Store that elements of arr1 in arr3    # that are present in hash    for i in range(n):         # If arr1 element is present        # in hash, store it in arr3        if (arr1[i] in hash):            arr3[k] = arr1[i]            k += 1                         del hash[arr1[i]]     # Copying 1st n elements of    # arr3 to arr1    for i in range(n):        arr1[i] = arr3[i] # Function to prarray elementsdef printArray(arr, n):         for i in arr:        print(i, end = " ")             print() # Driver Codeif __name__ == '__main__':         array1 = [ 7, 4, 8, 0, 1 ]    array2 = [ 9, 7, 2, 3, 6 ]    size = len(array1)         maximizeArray(array1, array2, size)    printArray(array1, size)     # This code is contributed by mohit kumar 29

C#

 // C# program to print the maximum elements// giving second array higher priorityusing System;using System.Collections.Generic; class GFG{ // Function to maximize array elementsstatic void maximizeArray(int[] arr1, int[] arr2){    // auxiliary array arr3 to store    // elements of arr1 & arr2    int []arr3 = new int;    for(int i = 0; i < arr3.Length; i++)    {        //arr2 has high priority        arr3[i] = 0;    }         // Arraylist to store n largest    // unique elements    List al = new List();         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(j);        }    }     int c = 0;    foreach(int x in al)    {        arr1[c++] = x;    }} // Function to print array elementsstatic void printArray(int[] arr){    foreach(int x in arr)    {        Console.Write(x + " ");    }} // Driver Codepublic 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 PrinciRaj1992

Javascript


Output:
9 7 6 4 8

Time complexity: O(n * log n).

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