What is the big O notation for this function?
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What is the big O notation for this function?
![def fibonacciRecursive (n):
if n<0:
print("Incorrect input")
elif n==0:
return 0
elif n=1:
return 1
else:
return fibonacciRecursive (n-1) +fibonacciRecursive (n-2)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F442cdeaf-d6f5-4c90-9eca-defd0049fe9f%2F1e12c7e6-a1ab-47de-bc09-53e4969cf13a%2F5rcbn0k_processed.png&w=3840&q=75)
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- int funcB(int); int funcA(int n) { if (n 4) { return n* funcA(n - 5); } else { return n- funcB(n - 1); int main() { cout << funcA(13); return 0; What is the output of this program? Please show our work.int FindSmallestVal() { int num = 0, min = 0; // reads num until the num > 0 while (num <= 0) { cin >> num; // finds the min value in the min,num min = num < min ? num : min; } // returns min return min; }int n = 1; int k - 2; int r = n; if (k < n) { r - k
- SYNTAX ERROR HELP - PYTHON This also happens for several other 'return result' lines in the code. import randomdef rollDice(): num1 = random.randint(1, 6) num2 = random.randint(1, 6) return num1, num2def determine_win_or_lose(num1, num2): result = -1 total = num1 + num2print(f"You rolled {num1} + {num2} = {total}")if total == 2 or total == 3 or total == 12: result = 0elif total == 7 or total == 11: result = 1else: print(f"point is {total}") x = determinePointValueResult(total)if x == 1: result = 1else: result = 0 return resultdef determinePointValueResult(pointValue): total = 0 result = -1while total != 7 and total != pointValue: num1, num2 = rollDice() total = num1 + num2if total == pointValue: result = 1elif total == 7: result = 0print(f"You rolled {num1} + {num2} = {total}")return resultwhile i < n: num1, num2 = rollDice() result = determine_win_or_lose(num1, num2)if result == 1: winCounter += 1 print("You…//Assignment 06 */public static void main[](String[] args) { String pass= "ICS 111"; System.out.printIn(valPassword(pass));} /* public static boolean valPassword(String password){ if(password.length() > 6) { if(checkPass(password) { return true; } else { return false; } }else System.out.print("Too small"); return false;} public static boolean checkPass (String password){ boolean hasNum=false; boolean hasCap = false; boolean hasLow = false; char c; for(int i = 0; i < password.length(); i++) { c = password.charAt(1); if(Character.isDigit(c)); { hasNum = true; } else if(Character.isUpperCase(c)) { hasCap = true; } else if(Character.isLowerCase(c)) { hasLow = true; } } return true; { return false; } }a) FindMinIterative public int FindMin(int[] arr) { int x = arr[0]; for(int i = 1; i < arr.Length; i++) { if(arr[i]< x) x = arr[i]; } return x; } b) FindMinRecursive public int FindMin(int[] arr, int length) { if(length == 1) return arr[0]; return Math.Min(arr[length - 1], Find(arr, length - 1)); } What is the Big-O for this functions. Could you explain the recurisive more in details ?
- #python def add_func(a,b): print(f"add_func output for 1 + 2: {add_func(1, 2)}") print(f"add_func output for good + day: {add_func('good',' day')}") (add_func(1,2) = 3, (add_func('good',' day') = "good day"Match the C-function on the left to the Intel assemble function on the right. W: cmpl $4 movl %edi , %edi jmp .L4(,%rdi,8) %edi .L3: movl $17, %eax ret .15: movl $3, %eax int A ( int x , int y) { int a ; if ( x == 0 ) else i f ( x == 1 ) a = 3 ; else i f ( x == 2 ) a = 2 0 ; else i f ( x == 3 ) a = 2 ; else i f ( x == 4 ) a = 1 ; ret .L6: a = 17; movl $20, %eax ret .L7: movl $2, %eax ret else a = 0; .L8: return a ; movl $1, %eax .L2: ret . section .rodata . L4: .quad .L3 .quad .L5 .quad .L6 .quad .L7 .quad .L8 X: testl %edi, %edi je cmpl je cmpl je стр1 je cmpl .L16 $1, %edi .L17 $2, %edi .L18 $3, %edi int B (int x, int y) { int a; switch (x) { .L19 $4, %edi %al movzbl %al, %eax case 0: a = 17; break; sete break; case 1: a = 3; case 2: a = 20; break; case 3: a = 2; break; case 4: a = 1; a = 0; } return a; ret .L16: break; movl $17, %eax ret .L17: movl $3, %eax } ret .L18: movl $20, %eax ret .L19: movl ret $2, %eax#include | int main() { int n=5; for (int i=1;i#include <stdio.h> int main(){ int arr[10]; int i; for (i=0; i<10; i++){ arr[i] = i; } for (i=0; i<10; i++){ printf("arr[%d] = %d\n", i, arr[i]); } return 0;} Copy the code and modify it as follows: Add another separate array of 10 integers. The following code parts are to be added (in the same order as specified) after the printing of the values of the first array:(1) Copy the content of the first array to the second array. Add 10 to each of the values of the second array. Make use of a single loop to achieve this part.(2) Print each element of the second array. Make use of another loop for this partLab Goal : This lab was designed to teach you more about recursion. Lab Description : Take a number and recursively determine how many of its digits are even. Return the count of the even digits in each number. % might prove useful to take the number apart digit by digit Sample Data : 453211145322224532714246813579 Sample Output : 23540CFG: Example 1 • Draw the CFG for the following code: int f(int n){ } int m = n* n; if (n < 0) else return 0; return m;SEE MORE QUESTIONS
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