Time complexity of merge sort
1 answer:
Answer:
The correct answer is "". A further explanation is given below.
Explanation:
- Throughout all the three instances (worst, average as well as best), the time complexity including its Merge sort seems to be
as the merge form often splits the array into two halves together tends to linear time to combine multiple halves.
- As an unsorted array, it needs an equivalent amount of unnecessary capacity. Therefore, large unsorted arrays are not appropriate for having to search.
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Answer:
<em>Python code is as follows: </em>
********************************************************************************
#function to get number up to any number of decimal places
def toFixed(value, digits):
return "%.*f" % (digits, value)
print("Enter the price: ", end='', flush=True) #prompt for the input of price
price = float(input()) #taken input
totalCost = price + 0.05 * price #calculating cost
print("Total Cost after the sales tax of 5% is applied: " + toFixed(totalCost,2)) #print and output totalCost
************************************************************************************
Answer:
See explaination
Explanation:
Code;
import java.util.Scanner;
public class NumberPattern {
public static int x, count;
public static void printNumPattern(int num1, int num2) {
if (num1 > 0 && x == 0) {
System.out.print(num1 + " ");
count++;
printNumPattern(num1 - num2, num2);
} else {
x = 1;
if (count >= 0) {
System.out.print(num1 + " ");
count--;
if (count < 0) {
System.exit(0);
}
printNumPattern(num1 + num2, num2);
}
}
}
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int num1;
int num2;
num1 = scnr.nextInt();
num2 = scnr.nextInt();
printNumPattern(num1, num2);
}
}
See attachment for sample output
