Skip to content
Related Articles

Related Articles

Save Article
Improve Article
Save Article
Like Article

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 <bits/stdc++.h>
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[0]);
    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<Integer> temp1 = new ArrayList<Integer>();
        ArrayList<Integer> temp2 = new ArrayList<Integer>();
        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<Integer,Integer> 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<Integer> res = new ArrayList<Integer>();
        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




<script>
// Make a set of maximum elements from two
// arrays A[] and B[]
function maximizeTheFirstArray(A, B, n)
{
 
    // Create copies of A[] and B[] and sort
    // the copies in descending order.
    let temp1 = new Array();
    let temp2 = new Array();
    for (let i of A)
    {
        temp1.push(i);
    }
    for (let i of B)
    {
        temp2.push(i);
    }
    temp1.sort((a, b) => a - b).reverse();
    temp2.sort((a, b) => a - b).reverse();
 
 
    // set maximum n distinct elements of
    // both sorted arrays in a map.
    let m = new Map();
    let i = 0, j = 0;
    while (m.size < n) {
        if (temp1[i] >= temp2[j]) {
            if (m.has(temp1[i])) {
                m.set(temp1[i], m.get(temp1[i]) + 1);
            }
            else {
                m.set(temp1[i], 1);
            }
            i++;
        }
        else {
            if (m.has(temp2[j])) {
                m.set(temp2[j], m.get(temp2[j]) + 1);
            }
            else {
                m.set(temp2[j], 1);
            }
            j++;
        }
    }
 
    // Copy elements of A[] to that
    // are present in hash m.
    let res = new Array();
    for (i = 0; i < n; i++)
        if (m.has(A[i]))
            res.push(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.has(B[i]) && m.get(B[i]) == 1)
            res.push(B[i]);
 
    // print result
    for (i = 0; i < n; i++)
        document.write(res[i] + " ");
}
 
// Driver program
let A = [9, 7, 2, 3, 6];
let B = [7, 4, 8, 0, 1];
let n = A.length;
maximizeTheFirstArray(A, B, n);
 
// This code is contributed by gfgking
 
</script>

Output: 

 9 7 6 4 8 

Time complexity: O(n Log n)




My Personal Notes arrow_drop_up
Recommended Articles
Page :