An array is manipulated k times so that each time the max value is divided by 2 and rounded up. I need to find its minimum sum after these k manipulations. k and all numbers in array num > 1. The method minSum receives an array called num and an integer k. The brute Python code that works for me with very bad time complexity is:
function minSum(arr, k) {
// Write your code here
let sum = 0;
while(k !==0){
let max = Math.max(...arr)
let index = arr.indexOf(max);
max = Math.ceil(max/2);
arr[index] = max;
k--;
}
sum = arr.reduce((a, b) => a + b, 0);
console.log(sum);
return sum;
}
Similar question relate to python is here. More efficient method of finding minimum sum after k operations
but nothing related to Javascript.
Here are the steps (using Java based on your first demand before the change to JavaScript):
public static int minSumJava_using_pqueue(int arr[], int k)
{
PriorityQueue<Integer> pq = new PriorityQueue<>(10, Collections.reverseOrder());
for (int val : arr) {
pq.add(val);
}
int new_val;
for(int i =0; i<k; i++)
{
new_val = pq.poll();
new_val = (int) Math.ceil(new_val/2.0);
pq.add(new_val);
}
int sum = 0;
for (Integer val: pq) {
sum += val;
}
return sum;
}
Check the source code:
public static void main(String[] args)
{
int k = 4;
int arr[] = {10,20,7};
int result = minSumJava_using_pqueue(arr, k);
System.out.println("min sum = "+ result);
}
The Result is indeed the same as in your example:
min sum = 14
Note: you can do exactly the same thing with JavaScript or any other programming language.