$\forall i,j \in \{1, \dots, n-1\}$ such that $i > j$, $(A[i + 1] - A[i]) > (A[j + 1] - A[j])$
Which one of the following gives the worst case time complexity of the fastest algorithm that can be designed for the problem?
Algorithm Condition Analysis
The problem requires checking a specific condition on an array $A$ of size $n$. The condition is stated as:
$ \forall i,j \in \{1, \dots, n-1\} \text{ such that } i > j, \quad (A[i + 1] - A[i]) > (A[j + 1] - A[j]) $
Let's define the difference between adjacent elements as $D[k] = A[k+1] - A[k]$ for $k \in \{1, \dots, n-1\}$.
The condition simplifies to ensuring that the sequence of differences, $D[1], D[2], \dots, D[n-1]$, is strictly increasing. That is:
$ D[1] < D[2] < \dots < D[n-1] $
Fastest Algorithm Verification
To verify this condition, the fastest algorithm must perform the following steps:
- Compute all adjacent differences $D[k] = A[k+1] - A[k]$ for $k$ from 1 to $n-1$. This requires accessing elements $A[1]$ through $A[n]$.
- Check if these computed differences are strictly increasing, i.e., $D[k] < D[k+1]$ for all $k$ from 1 to $n-2$.
Time Complexity Calculation
The process involves:
- Computing Differences: Calculating $n-1$ differences requires $n-1$ subtraction operations. Accessing the necessary array elements ($A[1]$ to $A[n]$) takes time proportional to $n$. This step is $\Theta(n)$.
- Checking Order: Comparing adjacent differences requires $n-2$ comparison operations. This step is also $\Theta(n)$.
The total time complexity is the sum of the time taken for these steps: $\Theta(n) + \Theta(n) = \Theta(n)$.
Since the algorithm must inspect all adjacent differences to guarantee the condition holds, it needs to access a significant portion of the array, establishing $\Omega(n)$ as a lower bound. As we have an algorithm that achieves $\Theta(n)$, this is the tightest bound for the fastest algorithm.
Conclusion
The worst-case time complexity of the fastest algorithm to check the given array condition is $\Theta(n)$.
Important Questions from Array
Which of the following expression will delete the entire array pointed to by q?
Consider the following C program segment.
#include <stdio.h>
int main()
{
char s1[7] = "1234";
char *p;p = s1 + 2;
*p = '0';printf("%s", s1);
return 0;
What will be printed by the program?
}#include <stdio.h> void foo(int *p, int x){ *p = x; } int main(){ int *z; int a = 20, b = 25; z = &a; foo(z, b); printf("%d", a); return 0; }The output of the given C program is __________. (Answer in integer)What is the output the given C language code snippet?
#include<stdio.h>
int main()
{
int x[] = {10,20,30};
int *p = x;
p++;
printf("%d",*p);
return 0;
}
