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 :

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// CPP program to print the maximum elements
// giving second array higher priority
#include <bits/stdc++.h>
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[0]);
    maximizeArray(array1, array2, size);
    printArray(array1, size);
}

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Output:

9 7 6 4 8

Time complexity: O(n * log n).



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