Question

int function3(int arr1[], int arr2[], int n) {    int count = 0; for (int i...

int function3(int arr1[], int arr2[], int n) {   
int count = 0;
for (int i = 1; i <= n / i && arr1[i] != arr2[i]; i++) {
count++;
}
return count;

find time complex
Engineering   Computer-Science   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

The function basically counts the total number of elements in array A which differ from the elements of array B at the same index.

Example: A = [1,2,3,4,5] and B = [2,2,3,4,6] , The answer for these two arrays will be 2. At index 0 the element differs and at index 4 the element differs.

Time Complexity: The for loop will run till n which is the total number of number of elements in the array. Rest of the operations are of unit 1 time complexity which are constant. So the overall time complexity will be O(n).

Happy Learning!

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Consider the following function: double arrayavg(int a[], int n){ int sum=0; for(int i=0; i!=n; ++i){ sum+=a[i];...
Consider the following function: double arrayavg(int a[], int n){ int sum=0; for(int i=0; i!=n; ++i){ sum+=a[i]; } return (double)sum/n; } Rewrite the function to eliminate the array subscripting (a[i]), using pointer arithmatic instead. Write a program that reads a line of input and checks if it is a palindrome (ignoring spaces) Write a program that takes the name of a file as a command line argument, and prints the contents of the file (similar to the linux 'cat' program). Write...
int algorithm ( int n ) {    if (n == 0)           return 1; else...
int algorithm ( int n ) {    if (n == 0)           return 1; else { return 2 * algorithm(n-1);    } }
1. public int add100(int[] array) { if (array.length < 100) { return 0; } int sum...
1. public int add100(int[] array) { if (array.length < 100) { return 0; } int sum = 0; for (int i = 0; i < 100; i++) { sum += array[i]; } return sum; } 2. int sum = 0; for (int i = 1; i <= n; i++) { for (int j = 0; j <= i; j++) { sum++; sum++; } } for (int i = 0; i < n; i += 2) { sum++; } 3. nt...
Given the following function: int bar(int n) {     if( n < 0 )         return...
Given the following function: int bar(int n) {     if( n < 0 )         return -2;     else if (n <= 1)         return 2;     else       return (3 + bar(n-1) + bar(n-2)); } What is returned by bar (4)? Show all the steps
For different input n, i.e., n = 1, n = 10, n = 20, write down...
For different input n, i.e., n = 1, n = 10, n = 20, write down the final value of counter for function1, function2, and function3. Explain the counter result through math calculation. #include <iostream> using namespace std; void function1(int n){ int count = 0; for(int x = 0; x < 12; x++){ cout<<"counter:"<< count++<<endl; } } void function2(int n){ int count = 0; for(int x = 0; x < n; x++){ cout<<"--- x="<<x<<"------"<<endl; for(int i =0; i < n;...
public int linearSearch2(ArrayList<Integer> pList, Integer pKey) { for (int i = 0; i <= pList.size() /...
public int linearSearch2(ArrayList<Integer> pList, Integer pKey) { for (int i = 0; i <= pList.size() / 2; ++i) { if (pList.get(i).equals(pKey)) { return i; } else if (pList.get(pList.size() - 1 - i).equals(pKey)) { return pList.size() - 1 - i; } } return -1; } ----- Q1)Which function f(n) counts how many times the key operation(s) is(are) performed as a function of the list size n when pKey is not in pList (the worst case behavior of linearSearch2() occurs when pKey...
Define a recurrence F(n) for the code: int a(int i, int j){ if( j <= 1)...
Define a recurrence F(n) for the code: int a(int i, int j){ if( j <= 1) return i+20; else return j + a(i+10, j-3); } This recurrence I've come up with is F(n) = 2 if j <= 1 and F(n) = 2 + F(i+10, j-3). Now I need to solve this recurrence and represent it in big O notation which I'm unsure how to do. The main thing throwing me off is that there are two variables in the...
In C can you change this code to use Tail Call Recursion int min (int n,...
In C can you change this code to use Tail Call Recursion int min (int n, int[] input) { int i; int result; for (result = input[0], i = 0; i < n; i += 1) { if (result > input[i]) result = input[i]; } return result; }
bool update(char userInput){ int i = board_height - 1; int j = userInput - 'A'; while(i>=0){...
bool update(char userInput){ int i = board_height - 1; int j = userInput - 'A'; while(i>=0){ if(board[i][j] == ' '){ board[i][j] = next; if(checkRow(i,j,next) || checkColumn(i,j,next) || checkLeftDiagonal(i,j,next) || checkRightDiagonal(i,j,next)){ global_state = next - '0'; } if(next == '1'){ next = '2'; }else{ next = '1'; } break; } i--; } if(i == -1){ return false; } return true; } How can this be turned into int update instead of bool update
What is the time complexity ? int fact(int n) { if( n==1) {return 1;} n=n*fact(n-1); return...
What is the time complexity ? int fact(int n) { if( n==1) {return 1;} n=n*fact(n-1); return n; }
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT