Compute for the time complexity (in milliseconds) of each letter of the given program segment below, then get the total time complexity. int main( ) { int a, b, c, d; for (a=0;a<10;a++) for (b=0;b<10;b++) for (c=0;c<10;c++) for (d=0;d<10;d++) { If ((a==0)&&(b==0)&&(c==0)&&(d==0)) { a=1, b=2,c=3,d=4; } cout<
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Compute for the time complexity (in milliseconds) of each letter of the given program segment below, then get the total time complexity.
- int main( )
- {
- int a, b, c, d;
- for (a=0;a<10;a++)
- for (b=0;b<10;b++)
- for (c=0;c<10;c++)
- for (d=0;d<10;d++)
- {
- If ((a==0)&&(b==0)&&(c==0)&&(d==0))
- {
- a=1, b=2,c=3,d=4;
- }
- cout<<a<<b<<c<<d;
- }
- }
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- Exercise 3. For each of the following program fragments give a (.) estimation of the running time as a function of n. (a) sum = 0; (b) (c) (d) (e) for (int i = 0; i< n * n; i++) { for(int j = 0; j < n/2; j++) sum++; } sum = 0; for (int i sum++; +; } for (int j } sum = for = } = 0; j < n/2; j++) { sum++; } 0; (int i = 0; i< n * n; i++) { for (int j = 0; j < n * n; j++) sum++ sum = 0; for (int i = 0; iDetermine the time complexity function of the program snippet below, int f1(int n) { if (n <= 1) return n; return 2 * f1(n/2); } int f2(int n) { if (n <= 1) return n; return f2(n/2) + f2(n/2); } please the answer must be in detail step by step until you get the time complexity function.Write a recursive function called print_num_pattern() to output the following number pattern. Given a positive integer as input (Ex: 12), subtract another positive integer (Ex: 3) continually until 0 or a negative value is reached, and then continually add the second integer until the first integer is again reached. For coding simplicity, output a space after every integer, including the last. Do not end output with a newline. Ex. If the input is: 12 3 the output is: 12 9 6 3 0 3 6 9 12 # TODO: Write recursive print_num_pattern() function if __name__ == "__main__": num1 = int(input()) num2 = int(input()) print_num_pattern(num1, num2)if (num1<=0): print(num1,end= '')Write a recursive algorithm with the following prototype: int divide (int x, int y); that returns x/y (integer division). You need not test for divide by 0. THE FUNCTION MUST BE RECURSIVE. (hint: base case should be when xWrite a complete C program to recursively find the largest element of a given array. Input Enter the size of the array : 8 Enter the array elements : 8, 3, 7, 1,9, 6, 2,4 Output 9Question 8 Consider the following code where n and m can be any number of more than 20. Select the correct time complexity. int iter_count = 0; for (int i = 0; i < n; i *= 2) { for (int j = 0; j < m; j++){ iter_count += 1; } for (int i = }) 0; iComputer scientists and mathematicians often use numbering systems other than base 10. Write a program that allows a user to enter a number and a base and then prints out the digits of the number in the new base. Use a recursive function baseConversion (num, base) to print the digits. Hint: Consider base 10. To get the rightmost digit of a base 10 number, simply look at the remainder after dividing by 10. For example, 153 % 10 is 3. To get the remaining digits, you repeat the process on 15, which is just 153 // 10. This same process works for any base. The only problem is that we get the digits in reverse order (right to left). The base case for the recursion occurs when num is less than base and the output is simply num. In the general case, the function (recursively) prints the digits of num // base and then prints num % base. You should put a space between successive outputs, since bases greater than 10 will print out with multi-character "digits." For example, baseConversion(1234,…Find the error in this C++ program and assure to submit a working program.#include <math.h> int adition (int num1, int num2); int subtration (int num1, int num2); int multiplication (int num1, int num2); int division (int num1, int num2); int remainder (int num1 int num2); int main(); { char option; int num1,num,result; float sqr; cout<<"\n\t\t\tSimple Calculator"<<endl; do { cout<<" Enter + for Addition\n Enter - for Subtraction\n Enter * for Multiplication\n Enter / for Division\n Enter R for Remainder\n Enter E to EXIT\n"; cout<<"\n\t\t\tEnter Option:"; cin>>option; if(option!='e'&&option!='E') { cout<<"\n\t\t\tEnter First Number:"; cin>>num1; cout<<"\n\t\t\tEnter Second Number:"; cin>num2; } if(option!='+'&&option!='-'&&option!='*'&&option!='/'&&option!='R'&&option!='e'&&option!='E') { cout<<"\n\t\t\tSELECT the VALID OPTION\n"; } else if(option=='+'); {…How can I apply this python code in the problem? def print_multiples(a,b,x): #Base Case/s #TODO: Add conditions here for your base case/s if True : return 0 #Recursive Case/s #TODO: Add conditions here for your recursive case/s else: return print_multiples(a, b, x) a = int(input("Please enter a: "))print(a)b = int(input("Please enter b: "))x = int(input("Please enter x: ")) print_multiples(a,b,x)1. Analyze the time complexity of the code segment and find their Big-O.void myfunction1(int n){for(int i=0; i < n; i++) {for(int j=0; j < n; j++) {for(int k=0; k < n; k++) {for(int m=0; m < n; m++) {printf("Hello!");} } } }}write a recursive function that checks see if the first letter matches the last letter, return the middle letters and check until only 0 or 1 letters are left. It returns True or False. Here is python code that needs to be fixed # Returns the first character of the string str def firstCharacter(str): return str[:1] # Returns the last character of a string str def lastCharacter(str): return str[-1:] # Returns the string that results from removing the first # and last characters from str def middleCharacters(str): return str[1:-1] def isPalindrome(str): # base case #1 # base case #2 # recursive case pass def isPalindrome(string): reversedString = string[::-1] if reversedString == string: return True else: return False inputString = input() string = inputString.replace(" ", "") if isPalindrome(string): print(inputString,"is a palindrome") else: print(inputString,"is not a palindrome") if __name__ ==…Recommended textbooks for youDatabase System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. 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