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Consider the program
void function(int n) {
int i, j, count=0;
for (i=n/2; i <= n; i++)
for (j = 1; j <= n; j = j*2)
count++;}
The complexity of the program is
void function(int n) {
int i, j, count=0;
for (i=n/2; i <= n; i++)
for (j = 1; j <= n; j = j*2)
count++;}
The complexity of the program is
- a)O(log n)
- b)O(n2)
- c)O(n2 log n)
- d)O(n log n)
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Consider the programvoid function(int n) {int i, j, count=0;for (i=n/2...
The outer loop runs n/2 times The inner loop runs logn times Therefore the total time complexity of the program is O(nlogn) which is option (D)
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Community Answer
Consider the programvoid function(int n) {int i, j, count=0;for (i=n/2...
This program calculates the number of prime numbers that are less than or equal to the input integer n.
The program starts by initializing three variables: i, j, and count. i is set to n/2, which means the program will start checking for prime numbers from the middle of the range (n/2) instead of the beginning (1). j is set to 2, which will be used to check if i is a prime number. count is initialized to 0, which will be used to keep track of the number of prime numbers found.
The program then enters a loop that will iterate from i down to 2 (inclusive). For each iteration, the program checks if i is a prime number by iterating from j up to the square root of i (rounded up to the nearest integer). If i is divisible by any number between 2 and the square root of i, then i is not a prime number and the loop moves on to the next iteration. If i is not divisible by any number between 2 and the square root of i, then i is a prime number and count is incremented.
Finally, the program prints out the value of count, which represents the number of prime numbers that are less than or equal to n.
The program starts by initializing three variables: i, j, and count. i is set to n/2, which means the program will start checking for prime numbers from the middle of the range (n/2) instead of the beginning (1). j is set to 2, which will be used to check if i is a prime number. count is initialized to 0, which will be used to keep track of the number of prime numbers found.
The program then enters a loop that will iterate from i down to 2 (inclusive). For each iteration, the program checks if i is a prime number by iterating from j up to the square root of i (rounded up to the nearest integer). If i is divisible by any number between 2 and the square root of i, then i is not a prime number and the loop moves on to the next iteration. If i is not divisible by any number between 2 and the square root of i, then i is a prime number and count is incremented.
Finally, the program prints out the value of count, which represents the number of prime numbers that are less than or equal to n.
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