# Maximize elements using another array

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 order ` `bool` `compare(``int` `a, ``int` `b) ` `{ ` `    ``return` `a > b; ` `} ` ` `  `// Function to maximize array elements ` `void` `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<``int``> 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 elements ` `void` `printArray(``int` `arr[], ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `        ``cout << arr[i] << ``" "``;     ` `    ``cout << endl; ` `} ` ` `  `// Driver Code ` `int` `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 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 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 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 `

## Python3

 `# Python3 program to print the maximum elements ` `# giving second array higher priority ` ` `  `# Function to maximize array elements ` `def` `maximizeArray(arr1, arr2, n): ` `     `  `    ``# Auxiliary array arr3 to store ` `    ``# elements of arr1 & arr2 ` `    ``arr3 ``=` `[``0``] ``*` `(``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 elements ` `def` `printArray(arr, n): ` `     `  `    ``for` `i ``in` `arr: ` `        ``print``(i, end ``=` `" "``) ` `         `  `    ``print``() ` ` `  `# Driver Code ` `if` `__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 priority ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `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``;  ` `    ``for``(``int` `i = 0; i < arr3.Length; i++) ` `    ``{ ` `        ``//arr2 has high priority ` `        ``arr3[i] = 0; ` `    ``} ` `     `  `    ``// Arraylist to store n largest ` `    ``// unique elements ` `    ``List<``int``> al = ``new` `List<``int``>(); ` `     `  `    ``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 elements ` `static` `void` `printArray(``int``[] arr) ` `{ ` `    ``foreach``(``int` `x ``in` `arr) ` `    ``{ ` `        ``Console.Write(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 PrinciRaj1992 `

Output:

```9 7 6 4 8
```

Time complexity: O(n * log 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 arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.