Solutions to Assessment

Transcription

1 Solutions to Assessment [1] What does the code segment below print? int fun(int x) ++x; int main() int x = 1; fun(x); printf( %d, x); return 0; Answer : 1. The argument to the function is passed by value. The changes to x in fun() will not affect the x value in the caller. [2] What is the output of the following code snippet? int fun(int x, int y, int z) y = y + 4; z = x + y + z; int main() int x = 10; int y = 3; fun(y, x, x); y = x + 2; fun(x, y, y); x = y + 2; printf( %d,%d, x, y); return 0; Answer : 14,12. The arguments x & y to the function are passed by value. The changes to x & y in fun() will not affect their values in the caller.

2 [3] What is the output of the following code snippet? int fun(int *x, int *y, int *z) *y = *y + 4; *z = *x + *y + *z; int main() int x = 10; int y = 3; fun(&y, &x, &x); y = x + 2; fun(&x, &y, &y); x = y + 2; printf( %d,%d, x, y); return 0; Answer : 107,105. The values x and y are passed by reference to the function. The changes done to x and y are reflected even after leaving fun(). [4] What does the below recursive function do? int *func(int *arr, int n) if(n == 1 n == 0) return arr; else int temp = arr[0]; arr[0] = arr[n - 1]; arr[n - 1] = temp; return func(++arr, n - 2); Answer : Reverses the given array. Each function call swaps the first and last element of the given array and calls the function recursively with an array of 1 st and last elements skipped.

3 [5] Let S(n) denote the sum of first n natural numbers. Specify which of the following statements are correct. S(n) = O(n 2 ) S(n) O(n 3 ) S(n) = O(n) S(n) = O(n 3 ) Answer : S(n) = O(n 2 ) and S(n) = O(n 3 ). The sum of first n natural numbers is n(n + 1) / 2, which is O(n 2 ) and can be written as O(n k ) for any k > 2. [6] Let S(n) denote the sum of squares of first n natural numbers. Specify which of the following statements are correct. S(n)=O(n 3 ) S(n) O(n 3 ) S(n)=O(n) S(n) O(n 2 ) Answer : S(n) = O(n 3 ) and S(n) O(n 2 ). The sum of first n natural numbers is n(n + 1)(2n + 1) / 6, which is O(n 3 ) but not O(n 2 ). [7] Consider the following segment of code: for(i = 0; i < n / 2; i++)... Let the body of for loop take O(n 3 ) time. Let S(n) represent the total running time of the whole segment. Specify which of the following statements are correct. S(n) = O(n 3 ) S(n) O(n 3 ) S(n) = O(n 4 ) S(n) O(n 5 ) Answer : S(n) = O(n 4 ) and S(n) O(n 3 ). The outer loop takes O(n) time. Hence the total complexity is O(n) times O(n 3 ) which is O(n 4 ) and is O(n k ) for any k < 4.

4 [8] Consider the following code segment. Assume n to be a positive integer. for(i = 1; i <= n; i++) for(j = 1; j <= n; j = j + i) printf( Hi ); Let T(n) represent the total running time of the above code segment. Specify which of the following statements are correct. T(n) = O(logn) T(n) = O(nlogn) T(n) = O(n 2 ) T(n) O(n 4 ) Answer : T(n) = O(nlogn) and T(n) = O(n 2 ). The complexity is O(n) times O(logn), hence O(nlogn), and this implies it is also O(n 2 ). [9] Let the function f(n) = n + 2n + 3n+ + (n 1 )*n. Assume n to be a positive integer. Specify whether the following statement is true or false. f(n) = O(n 2 ). Answer : False. f(n) = n * ( (n - 1)) = n * n * (n - 1) / 2 = O(n 3 ). [10] Specify whether the following statement is true or false. If a function T(n) = O(n i ), then T(n) = O(n j ) whenever i < j and i, j are positive integers. Answer : True. O(n i ) can be written as O(n j ) for any j > i. Example O(n 2 ) = O(n 3 ). But the converse is not true. Example O(n 3 ) cannot be written as O(n 2 ).

5 Solutions to Programming Assignments Question 1: Count pairs such that x y > y x Given a sequence A of N positive integers, write a program to find the number of pairs (A[i], A[j]) such that i < j and A[i] A[j] > A[j] A[i] (A[i] raised to the power A[j] > A[j] raised to the power A[i]). You are provided with a function named power( ) that takes two positive integers x & y and returns x y. If y is 0, the function returns 1. The prototype of this function is int power(int x, int y); You do not have to write the program for power ( ) function. This function is automatically added at the end of the code segment that you write. In your program, use this function to count the number of such pairs. INPUT: Line 1 contains N. Line 2 contains a sequence of N integers, separated by whitespace. OUTPUT: A single integer representing the number of required pairs. CONSTRAINTS: The inputs will satisfy the following properties. It is not necessary to validate the inputs. 1 <= N <= 30 0 <= A[i] <= 8 The input sequence can have repetitions and the count must reflect that. Public test cases: Input

6 Private test cases: Input Solution: #include <stdio.h> int power(int x, int y); int main() int i, j, n; int a[100]; scanf("%d", &n); for(i = 0; i < n; i++) scanf("%d", &a[i]); int count = 0;

7 for(i = 0; i < n; i++) for(j = i + 1; j < n; j++) if(power(a[i], a[j]) > power(a[j], a[i])) count++; printf("%d", count); return 0; Explanation: The solution above is based on the following approach, For each element a[i], count the number of elements to the right of a[i] such that a[i] a[j] > a[j] a[i]. i.e. for each a[i], scan through a[i + 1], a[i + 2],, so on till a[n] and increment the answer if the property is satisfied. The powers can be calculated using the call to power() function.

8 Question 2: Evaluating a Polynomial You are given a polynomial of degree n. The polynomial is of the form P(x) = a n x n + a n- 1x n a 0, where the a i s are the coefficients. Given an integer x, write a program that will evaluate P(x). You are provided with a function named power( ) that takes two positive integers x & y and returns x y. If y is 0, the function returns 1. The prototype of this function is int power(int x, int y); You do not have to write the program for power ( ) function. This function is automatically added at the end of the code segment that you write. INPUT: Line 1 contains the integers n and x separated by whitespace. Line 2 contains the coefficients a n, a n-1, a 0 separated by whitespace. OUTPUT: A single integer which is P(x). CONSTRAINTS: The inputs will satisfy the following properties. It is not necessary to validate the inputs. 1 <= n <= 10 1 <= x <= 10 0 <= a i <=10 Public test cases: Input Private test cases:

9 Input Solution: #include <stdio.h> int power(int x, int y); int main() int i, n, x; int a[100]; int sum = 0; scanf("%d %d", &n, &x); for(i = 0; i <= n; i++) scanf("%d", &a[i]); for(i = 0; i <= n; i++) sum += a[i] * power(x, n - i);

10 printf("%d",sum); return 0; Explanation: The sum can be calculated by adding up the terms (a[i] * x n i ) for i = 0, 1,, upto n 1 and the a[i] s are the coefficients of the polynomial as given in the input format. a[0] is the coefficient of x n, a[1] is the coefficient of x n-1 and so on. For calculating x n i, the power() function can be used.

11 Question 3: Evaluating a Recurrence Relation A recurrence relation T is defined on n >= 0 and is given as T(n) = T(n-1) + 2*n with the base case T(0) = 1. You will be given one integer k and you have to write a program to find out T(k). The program must implement T( ) recursively. INPUT: One integer k. OUTPUT: T(k) CONSTRAINTS: The input k will satisfy the constraints 0 <= k <= You do not have to check if the input given to you falls outside this range. Public test cases: Input Private test cases: Input Solution: #include <stdio.h> int T(int k);

12 int main() int k; scanf("%d", &k); printf("%d", T(k)); return 0; int T(int k) if(k == 0) return 1; // Base condition. else return T(k - 1) + 2 * k; Explanation: The above solution is a recursive implementation of the given recurrence. The base condition given is T(0) = 1 and this is the first check in the above code. If the k value is greater than zero the function makes a recursive call to T(k - 1) and adds 2 * k to it. For all values of k >= 0, the function is guaranteed to terminate.

13 Question 4: Palindrome Checker Write a program which takes a string as input and prints the longest prefix of the string whose reverse is a valid suffix of the string. For example, if "notation" is given as input, the prefix "no" is the longest prefix that has a matching valid suffix (namely "on") at the end. If a string is a palindrome (i.e. if the string is the same as its reverse), then the prefix is the whole string itself. For example, if input is "civic", then the longest prefix is "civic" itself. If the string has no valid prefix, print 0. For example, if the input string is "bird", there is a mismatch at the first character itself and there is no valid prefix. The symbols in the string will only be from the set A-Z,a-z. The symbols are case sensitive. i.e. A and a are considered to be different. You are provided with a function called printchars ( ) that prints a character array from the starting position pointed to by p to the ending position pointed to by q. If p is NULL, it prints 0. The prototype of the function is: void printchars(char *p, char *q); You do not have to write the program for printchars ( ) function. This function is automatically added at the end of the code segment that you write. You can use string functions in the library if you desire to. INPUT: Input is a string of length N composed of symbols only from the set A-Za,z. OUTPUT: The longest prefix that has a corresponding matching suffix and 0 if such a prefix does not exist. CONSTRAINTS: The inputs will satisfy the following properties. It is not necessary to validate the inputs. 1<=N<=99 Public Test Cases:

14 Input notation b no b camera 0 malayalam racecar mal racecar Private Test Cases: Input murdrum murdrum catalyst 0 v MadAM civic aa v M civic aa ab 0 Solution: Please watch the video at