The following algorithm
Count the number of primitive operations (evaluating an expression, indexing into an array). What is the time complexity of this algorithm in big-O notation?
for all i = n down to 1 do
for all j = n down to i do
print A[i] A[j]
end for
end for
Answer:
statement | # primitive operations |
---|---|
for all i = n down to 1 do | \( n+(n+1) \) |
for all j = n down to i do | \( 3+5+\cdots+(2n+1) = n(n+2) \) |
print A[i] A[j] | \( (1+2+\cdots+n)\cdot 2 = n(n+1) \) |
end for | |
end for |
This totals to \( 2n^2+5n+1 \), which is \( O(n^2) \)
Develop an algorithm to determine
if a character array of length n
encodes a palindrome —
that is, it reads the same forward and backward.
For example, racecar
is a palindrome.
Write the algorithm in pseudocode.
Analyse the time complexity of your algorithm.
Implement your algorithm in C. Your program should accept a single command line argument, and check whether it is a palindrome. Examples of the program executing are
./palindrome racecar yes ./palindrome reviewer no
Hint:
You may use the standard library function strlen(3),
which has prototype size_t strlen(char *)
,
is defined in <string.h>
,
and which computes the length of a string
(without counting its terminating '\0'
-character).
Answer:
isPalindrome(A): Input array A[0..n-1] of chars Output true if A palindrome, false otherwise i = 0; j = n-1 while i < j do if A[i] ≠ A[j] then return false end if i = i + 1; j = j - 1 end while return true
Time complexity analysis: There are at most \( \lfloor n/2 \rfloor \) iterations of the loop, and the operations inside the loop are \( O(1) \). Hence, this algorithm takes \( O(n) \) time.
#include <stdbool.h> #include <stdio.h> #include <string.h> static bool isPalindrome (char A[], int len); int main (int argc, char *argv[]) { if (argc == 2) { if (isPalindrome (argv[1], strlen (argv[1]))) printf ("yes "); else printf ("no "); } return 0; } static bool isPalindrome (char A[], int len) { int i = 0; int j = len - 1; while (i < j) { if (A[i] != A[j]) return true; i++; j--; } return false; }
Let \( p(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \) be a polynomial of degree \( n \). Design an \( O(n) \)-time algorithm for computing \( p(x) \).
Hint: Assume that the coefficients \( a_i \) are stored in an array A[0..n].
Answer:
Rewriting \( p(x) \) as \( ( \cdots ((a_n x + a_{n-1}) x + a_{n-2}) x + \cdots + a_1) x + a_0 \) leads to the following algorithm:evalPolynomial(A,n,x): p = A[n] for all i = n-1 down to 0 do p = p * x + A[i] end for
This is \( O(n) \).
A vector \( V \) is called sparse if most of its elements are 0.
In order to store sparse vectors efficiently,
we can use a list L
to store only its non-zero elements.
Specifically, for each non-zero element \( V[i] \),
we store an index-value pair \( (i, V[i]) \) in L
.
We call L
the compact form of \( V \).
For example,
the 8-dimensional vector
\( V=(2.3,0,0,0,-5.61,0,0,1.8) \)
could be stored in a list L
of size 3:
\( [ (0,2.3), (4, -5.61), (7, 1.8) ] \).
Describe an efficient algorithm for adding two sparse vectors \( V_1 \) and \( V_2 \) of equal dimension, but given in compact form. The result should be in compact form too. What is the time complexity of your algorithm, depending on the sizes \( m \) and \( n \) of the compact forms of \( V_1 \) and \( V_2 \), respectively?
Hint: The sum of two vectors \( V_1 \) and \( V_2 \) is defined as usual; for example, (2.3,-0.1,0,0,1.7,0,0,0) + (0,3.14,0,0,-1.7,0,0,-1.8) = (2.3,3.04,0,0,0,0,0,-1.8).
Answer:
In the algorithm below,
L[i].x
denotes the first component (the index) of a pair L[i]
,
and L[i].v
denotes its second component (the value).
VectorSum(L1,L2): Input compact forms L1 of length m and L2 of length n Output compact form of L1+L2 i = 0 // index into L1 j = 0 // index into L2 k = 0 // index into L3 while i ≤ m-1 ∧ j ≤ n-1 do if L1[i].x = L2[j].x then // found index with entries in both vectors if L1[i].v ≠ -L2[j].v then // only add if the values don't add up to 0 L3[k] = (L1[i].x, L1[i].v + L2[j].v) k = k + 1 end if i = i + 1; j = j + 1 else if L1[i].x < L2[j].x then // copy an entry from L1 L3[k] = (L1[i].x, L1[i].v) i = i + 1; k = k + 1 else // copy an entry from L2 L3[k] = (L2[j].x, L2[j].v) j = j + 1; k = k + 1 end if end while // copy any remaining pairs from L1 while i < m-1 do L3[k] = (L1[i].x, L1[i].v) i = i + 1; k = k + 1 end while // copy any remaining pairs from L2 while j < n-1 do L3[k] = (L2[j].x, L2[j].v) j = j + 1; k = k + 1 end while
Time complexity analysis:
Adding a pair to L3
takes \( O(1) \) time.
At most \( m + n \) pairs
from L1
and L2
are added.
Therefore, the time complexity is \( O(m + n) \).
Suppose that you are given two stacks of non-negative integers \( A \) and \( B \) and a target threshold \( k \ge 0 \). Your task is to determine the maximum number of elements that you can pop from \( A \) and \( B \) so that the sum of these elements does not exceed \( k \).
For example, given the stacks
The maximum number of elements that can be popped without exceeding \( k = 10 \) is 4:
If \( k = 7 \), then the answer would be 3: the top element of A and the top two elements of B.
Write an algorithm (in pseudocode) to determine this maximum for any given stacks \( A \) and \( B \) and threshold \( k \). As usual, the only operations you can perform on the stacks are pop and push. You are permitted to use a third "helper" stack, but no other aggregate data structure.
Determine the time complexity of your algorithm depending on the sizes \( m \) and \( n \) of input stacks \( A \) and \( B \).
Answer:
MaxElementsToPop(A,B,k): Input stacks A and B, target threshold k ≥ 0 Output maximum number of elements that can be popped from A and B so that their sum does not exceed k sum = 0; count = 0; C = new Stack // Phase 1: Determine how many elements can be popped just from A, // and push those onto the helper stack C while sum ≤ k ∧ stack A not empty do v = pop(A); push(C, v) sum = sum + v; count = count + 1 end while // have we exceeded k? if sum > k then // subtract last element that's been popped off A; // this is the best you can do with elements from A only. v = pop(C); sum = sum - v; count = count - 1 end if best = count // Phase 2: add one element from B at a time; whenever the threshold // is exceeded, subtract more elements originally from A (now in C), // to get back below threshold while stack B not empty do v = pop(B); sum = sum + v; count = count + 1 while sum > k ∧ stack C not empty do v = pop(C); sum = sum - v; count = count - 1 end while // update each time you got a better score if sum ≤ k ∧ count > best then best = count end if end while return best
Time complexity analysis: In phase 1, the worst case is when all \( m \) elements in stack A need to be visited, for a maximum of \( m + 1 \) pop, and \( m \) push operations. In phase 2, the worst case is when all elements have to be taken from both B and C, for a maximum of \( m + n \) pop operations. Assuming that push and pop take constant time, the overall complexity is \( O(m+n) \). This makes it a linear-time solution.