Javascript Program for Maximize elements using another array
Last Updated :
24 Mar, 2023
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 :
Javascript
<script>
function maximizeArray(arr1,arr2)
{
let arr3 = new Array(10);
for (let i = 0; i < arr3.length; i++)
{
arr3[i] = 0;
}
let al = [];
for (let i = 0; i < arr2.length; i++)
{
if (arr3[arr2[i]] == 0)
{
arr3[arr2[i]] = 2;
al.push(arr2[i]);
}
}
for (let i = 0; i < arr1.length; i++)
{
if (arr3[arr1[i]] == 0)
{
arr3[arr1[i]] = 1;
al.push(arr1[i]);
}
}
let count = 0;
for (let j = 9; j >= 0; j--)
{
if (count < arr1.length &
(arr3[j] == 2 || arr3[j] == 1))
{
count++;
}
else
{
if (al.indexOf(j)>0)
al.splice(al.indexOf(j),1);
}
}
let i = 0;
for (let x = 0; x < al.length; x++)
{
arr1[i++] = al[x];
}
}
function printArray(arr)
{
for (let x=0; x<arr.length;x++)
{
document.write(arr[x] + " " );
}
}
let arr1=[7, 4, 8, 0, 1];
let arr2=[9, 7, 2, 3, 6];
maximizeArray(arr1,arr2);
printArray(arr1);
</script>
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Time complexity: O(n * log n).
Space Complexity: O(n) as list has been created.
Please refer complete article on Maximize elements using another array for more details!
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