# Maximum array from two given arrays keeping order same

• Difficulty Level : Medium
• Last Updated : 22 Jun, 2021

Given two same-sized arrays A[] and B[] (both arrays contain distinct elements individually but may have some common elements), the task is to form a third (or result) array of the same size. The resulting array should have maximum n elements from both arrays. It should have chosen elements of A[] first, then chosen elements of B[] in the same order as they appear in original arrays. If there are common elements, then only one element should be present in res[] and priority should be given to A[].

Examples:

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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

```Input :  A[] =  [ 9 7 2 3 6 ]
B[] =  [ 7 4 8 0 1 ]
Output : res[] = [9 7 6 4 8]
res[] has maximum n elements of both A[]
and B[] such that elements of A[] appear
first (in same order), then elements of B[].
Also 7 is common and priority is given to
A's 7.

Input :  A[] = [ 6 7 5 3 ]
B[] = [ 5 6 2 9 ]
Output : res[] = [ 6 7 5 9 ]```

1) Create copies of both arrays and sort the copies in decreasing order.
2) Use a hash to pick unique n maximum elements of both arrays, giving priority to A[].
3) Initialize result array as empty.
4) Traverse through A[], copy those elements of A[] that are present in the hash. This is done to keep the order of elements the same.
5) Repeat step 4 for B[]. This time we only consider those elements that are not present in A[] (Do not appear twice in the hash).

Below is the implementation of the above idea.

## C++

 `// Make a set of maximum elements from two``// arrays A[] and B[]``#include ``using` `namespace` `std;` `void` `maximizeTheFirstArray(``int` `A[], ``int` `B[],``                                    ``int` `n)``{``    ``// Create copies of A[] and B[] and sort``    ``// the copies in descending order.``    ``vector<``int``> temp1(A, A+n);``    ``vector<``int``> temp2(B, B+n);``    ``sort(temp1.begin(), temp1.end(), greater<``int``>());``    ``sort(temp2.begin(), temp2.end(), greater<``int``>());` `    ``// Put maximum n distinct elements of``    ``// both sorted arrays in a map.``    ``unordered_map<``int``, ``int``> m;``    ``int` `i = 0, j = 0;``    ``while` `(m.size() < n)``    ``{``         ``if` `(temp1[i] >= temp2[j])``         ``{``            ``m[temp1[i]]++;``            ``i++;``         ``}``         ``else``         ``{``            ``m[temp2[j]]++;``            ``j++;``         ``}``    ``}` `    ``// Copy elements of A[] to that``    ``// are present in hash m.``    ``vector<``int``> res;``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `(m.find(A[i]) != m.end())``           ``res.push_back(A[i]);` `    ``// Copy elements of B[] to that``    ``// are present in hash m. This time``    ``// we also check if the element did``    ``// not appear twice.``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `(m.find(B[i]) != m.end() &&``            ``m[B[i]] == 1)``           ``res.push_back(B[i]);` `    ``// print result``    ``for` `(``int` `i = 0; i < n; i++)``        ``cout << res[i] << ``" "``;``}` `// driver program``int` `main()``{``    ``int` `A[] = { 9, 7, 2, 3, 6 };``    ``int` `B[] = { 7, 4, 8, 0, 1 };``    ``int` `n = ``sizeof``(A) / ``sizeof``(A);``    ``maximizeTheFirstArray(A, B, n);``    ``return` `0;``}`

## Java

 `// Make a set of maximum elements from two``// arrays A[] and B[]``import` `java.io.*;``import` `java.util.*;``class` `GFG``{``    ` `    ``static` `void` `maximizeTheFirstArray(``int``[] A, ``int``[] B,``int` `n)``    ``{``      ` `        ``// Create copies of A[] and B[] and sort``        ``// the copies in descending order.``        ``ArrayList temp1 = ``new` `ArrayList();``        ``ArrayList temp2 = ``new` `ArrayList();``        ``for``(``int` `i : A)``        ``{``            ``temp1.add(i);``        ``}``        ``for``(``int` `i:B)``        ``{``            ``temp2.add(i);``        ``}``        ``Collections.sort(temp1, Collections.reverseOrder());``        ``Collections.sort(temp2, Collections.reverseOrder());``        ` `        ``// Put maximum n distinct elements of``        ``// both sorted arrays in a map.``        ``Map m = ``new` `HashMap<>();``        ``int` `i = ``0``, j = ``0``;``        ``while` `(m.size() < n)``        ``{``             ``if` `(temp1.get(i) >= temp2.get(j))``             ``{``                ``if``(m.containsKey(temp1.get(i)))``                ``{``                    ``m.put(temp1.get(i), m.get(temp1.get(i)) + ``1``);``                ``}``                ``else``                ``{``                    ``m.put(temp1.get(i), ``1``);``                ``}``                ``i++;``             ``}``             ``else``             ``{``                ``if``(m.containsKey(temp2.get(j)))``                ``{``                    ``m.put(temp2.get(j), m.get(temp2.get(j)) + ``1``);``                ``}``                ``else``                ``{``                    ``m.put(temp2.get(j), ``1``);``                ``}``                ``j++;``             ``}``        ``}``        ` `        ``// Copy elements of A[] to that``        ``// are present in hash m.``        ``ArrayList res = ``new` `ArrayList();``        ``for` `(i = ``0``; i < n; i++)``            ``if` `(m.containsKey(A[i]))``               ``res.add(A[i]);``     ` `        ``// Copy elements of B[] to that``        ``// are present in hash m. This time``        ``// we also check if the element did``        ``// not appear twice.``        ``for` `(i = ``0``; i < n; i++)``            ``if` `(m.containsKey(B[i]) && m.get(B[i]) == ``1``)``               ``res.add(B[i]);``     ` `        ``// print result``        ``for` `(i = ``0``; i < n; i++)``            ``System.out.print(res.get(i)+``" "``);``    ``}``     ` `    ``// Driver program   ``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `A[] = { ``9``, ``7``, ``2``, ``3``, ``6` `};``        ``int` `B[] = { ``7``, ``4``, ``8``, ``0``, ``1` `};``        ``int` `n = A.length;``        ``maximizeTheFirstArray(A, B, n);``    ``}``}` `// This code is contributed by rag2127`

## Python3

 `# Python3 program to implement the``# above approach``# Make a set of maximum elements``# from two arrays A[] and B[]``from` `collections ``import` `defaultdict` `def` `maximizeTheFirstArray(A, B, n):` `    ``# Create copies of A[] and B[]``    ``# and sort the copies in``    ``# descending order.``    ``temp1 ``=` `A.copy()``    ``temp2 ``=` `B.copy()``    ``temp1.sort(reverse ``=` `True``)``    ``temp2.sort(reverse ``=` `True``)` `    ``# Put maximum n distinct``    ``# elements of both sorted``    ``# arrays in a map.``    ``m ``=` `defaultdict(``int``)``    ``i ``=` `0``    ``j ``=` `0``;``    ` `    ``while` `(``len``(m) < n):``         ``if` `(temp1[i] >``=` `temp2[j]):``            ``m[temp1[i]] ``+``=` `1``            ``i ``+``=` `1`       `         ``else``:``            ``m[temp2[j]] ``+``=` `1``            ``j ``+``=` `1` `    ``# Copy elements of A[] to that``    ``# are present in hash m.``    ``res ``=` `[]``    ` `    ``for` `i ``in` `range` `(n):``        ``if` `(A[i] ``in` `m):``           ``res.append(A[i])` `    ``# Copy elements of B[] to that``    ``# are present in hash m. This time``    ``# we also check if the element did``    ``# not appear twice.``    ``for` `i ``in` `range` `(n):``        ``if` `(B[i] ``in` `m ``and``            ``m[B[i]] ``=``=` `1``):``           ``res.append(B[i])` `    ``# Print result``    ``for` `i ``in` `range` `(n):``        ``print` `(res[i], end ``=` `" "``)` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``  ` `    ``A ``=` `[``9``, ``7``, ``2``, ``3``, ``6``]``    ``B ``=` `[``7``, ``4``, ``8``, ``0``, ``1``]``    ``n ``=` `len``(A)``    ``maximizeTheFirstArray(A, B, n);``  ` `# This code is contributed by Chitranayal`

## C#

 `// Make a set of maximum elements from two``// arrays A[] and B[]``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `static` `void` `maximizeTheFirstArray(``int``[] A, ``int``[] B,``                                  ``int` `n)``{``    ` `    ``// Create copies of A[] and B[] and sort``    ``// the copies in descending order.``    ``List<``int``> temp1 = ``new` `List<``int``>();``    ``List<``int``> temp2 = ``new` `List<``int``>();``    ` `    ``foreach``(``int` `i ``in` `A)``    ``{``        ``temp1.Add(i);``    ``}``    ` `    ``foreach``(``int` `i ``in` `B)``    ``{``        ``temp2.Add(i);``    ``}``    ` `    ``temp1.Sort();``    ``temp1.Reverse();``    ``temp2.Sort();``    ``temp2.Reverse();``     ` `    ``// Put maximum n distinct elements of``    ``// both sorted arrays in a map.``    ``Dictionary<``int``,``               ``int``> m = ``new` `Dictionary<``int``,``                                       ``int``>();``    ``int` `I = 0, j = 0;``    ``while` `(m.Count < n)``    ``{``        ``if` `(temp1[I] >= temp2[j])``        ``{``            ``if` `(m.ContainsKey(temp1[I]))``            ``{``                ``m[temp1[I]]++;``            ``}``            ``else``            ``{``                ``m.Add(temp1[I], 1);``            ``}``            ``I++;``        ``}``        ``else``        ``{``            ``if` `(m.ContainsKey(temp2[j]))``            ``{``                ``m[temp2[j]]++;``            ``}``            ``else``            ``{``                ``m.Add(temp2[j], 1);``            ``}``            ``j++;``        ``}``    ``}` `    ``// Copy elements of A[] to that``    ``// are present in hash m.``    ``List<``int``> res = ``new` `List<``int``>();``    ``for``(``int` `i = 0; i < n; i++)``        ``if` `(m.ContainsKey(A[i]))``            ``res.Add(A[i]);``  ` `    ``// Copy elements of B[] to that``    ``// are present in hash m. This time``    ``// we also check if the element did``    ``// not appear twice.``    ``for``(``int` `i = 0; i < n; i++)``        ``if` `(m.ContainsKey(B[i]) && m[B[i]] == 1)``            ``res.Add(B[i]);``  ` `    ``// print result``    ``for``(``int` `i = 0; i < n; i++)``        ``Console.Write(res[i] + ``" "``);``}``  ` `// Driver Code``static` `public` `void` `Main()``{``    ``int``[] A = { 9, 7, 2, 3, 6 };``    ``int``[] B = { 7, 4, 8, 0, 1 };``    ``int` `n = A.Length;``    ` `    ``maximizeTheFirstArray(A, B, n);``}``}` `// This code is contributed by avanitrachhadiya2155`

## Javascript

 ``

Output:

` 9 7 6 4 8 `

Time complexity: O(n Log n)

My Personal Notes arrow_drop_up