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>// Javascript program to print the maximum elements// giving second array higher priority// Function to maximize array elementsfunction maximizeArray(arr1,arr2){ // auxiliary array arr3 to store // elements of arr1 & arr2 let arr3 = new Array(10); for(let i = 0; i < arr3.length; i++) { // arr2 has high priority arr3[i] = 0; } // Arraylist to store n largest // unique elements let al = []; for(let 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.push(arr2[i]); } } for(let 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.push(arr1[i]); } } // to get only highest n elements(arr2+arr1) // and remove others from arraylist let count = 0; for(let 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 { 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 to print array elementsfunction printArray(arr){ for(let x=0; x<arr.length;x++) { document.write(arr[x] + " "); }}// Driver Codelet arr1=[7, 4, 8, 0, 1];let arr2=[9, 7, 2, 3, 6];maximizeArray(arr1,arr2);printArray(arr1); // This code is contributed by patel2127</script> |
9 7 6 4 8
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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