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13n+ 10 = O(n²) is also:
O
ОО
O
3
O (little o)
Ω

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1. Both mergesort and quicksort uses divide and conquer paradigm to sort unsorted list. (a) Imagine you want to write the quicksort algorithm to sort an array into non-increasing order. Write down the partition algorithm that is used in the divide phase in the quicksort algorithm. Show that the time complexity of this algorithm is 0(n). (b) Identify the worst case situation in the naive quicksort algorithm and show that in the worst case situation its time complexity is O(n2)
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Write code for. Computes the smallest x that satisfies the chinese remainder theorem for a system of equations. The system of equations has the form: x % nums[0] = rems[0] x % nums[1] = rems[1] ... x % nums[k - 1] = rems[k - 1] Where k is the number of elements in nums and rems, k > 0. All numbers in nums needs to be pariwise coprime otherwise an exception is raised returns x: the smallest value for x that satisfies the system of equations.
There is more than one way to calculate the value of T. One way that this can be done is by generating random numbers. This works by recognizing that if you take a unit square that you can draw a quarter circle of unit radius inside the square. The area of the quarter circle is exactly π/4 and the area of the square is 1. So if you randomly generate a pair of uniform numbers between 0 and 1 they will be distributed uniformly across the square. If you count the total number of points generated and the number of points (x, y) where x² + y² = ² < 1 then the ratio of those two numbers will tend towards the ratios of the areas of the square and the circle as the number of points generated increases. The ratio of the areas is just π/4 so if you take that ratio and multiply it by 4 you get an estimate for . a) b Write code to estimate π using this method. You can generate random numbers in the range 0
Given a positive integer 'n', find and return the minimum number of steps that 'n' has to take to get reduced to 1. You can perform any one of the following 3 steps:1.) Subtract 1 from it. (n = n - 1) ,2.) If its divisible by 2, divide by 2.( if n % 2 == 0, then n = n / 2 ) ,3.) If its divisible by 3, divide by 3. (if n % 3 == 0, then n = n / 3 ). Write brute-force recursive solution for this.Input format :The first and the only line of input contains an integer value, 'n'.Output format :Print the minimum number of steps.Constraints :1 <= n <= 200 Time Limit: 1 secSample Input 1 :4Sample Output 1 :2 Explanation of Sample Output 1 :For n = 4Step 1 : n = 4 / 2 = 2Step 2 : n = 2 / 2 = 1 Sample Input 2 :7Sample Output 2 :3Explanation of Sample Output 2 :For n = 7Step 1 : n = 7 - 1 = 6Step 2 : n = 6 / 3 = 2 Step 3 : n = 2 / 2 = 1 SolutionDp///.
bool isprime(long n) /* fixed from to https://www.geeksforgeeks.org/euclid-euler-theorem/?ref=lbp */{ // check whether a number is prime or not int i; for (i = 2; i * i <= n; i++) if (n % i == 0) return false; return true;}
Given a positive integer 'n', find and return the minimum number of steps that 'n' has to take to get reduced to 1. You can perform any one of the following 3 steps:1.) Subtract 1 from it. (n = n - 1) ,2.) If its divisible by 2, divide by 2.( if n % 2 == 0, then n = n / 2 ) ,3.) If its divisible by 3, divide by 3. (if n % 3 == 0, then n = n / 3 ). Write brute-force recursive solution for this.Input format :The first and the only line of input contains an integer value, 'n'.Output format :Print the minimum number of steps.Constraints :1 <= n <= 200 Time Limit: 1 secSample Input 1 :4Sample Output 1 :2 Explanation of Sample Output 1 :For n = 4Step 1 : n = 4 / 2 = 2Step 2 : n = 2 / 2 = 1 Sample Input 2 :7Sample Output 2 :3Explanation of Sample Output 2 :For n = 7Step 1 : n = 7 - 1 = 6Step 2 : n = 6 / 3 = 2 Step 3 : n = 2 / 2 = 1 SolutionDp///.
Bum 0; for (int i = 1; i
The statement 2n< n! is valid for a.All positive integers b.All positive integers n ≥ 4 c.For only n=4d.All positive integers n ≤ 4
Part 3: Negative Binary Numbers This standard method of binary number representation doesn't account for negative numbers. We can't just throw a negative sign in front of a binary number though - everything in a computer must be represented by 1's and 0's. One way we can get around this is by using the very first digit of each binary number to represent the sign: 0 for positive and 1 for negative. This is called 1's complement representation. 00010110 -> 0 0010110 -> positive 22 10010110 -> 1 0010110 -> negative 22 Using 1's complement representation we can represent the numbers -127 through 127 in a single byte (note that we can only go half as high as an unsigned number because half of the possible representations for that byte are being used for the negative numbers). However, there is a flaw with 1's complement: the representation of the number 0. 00000000 -> 0 0000000 -> zero 10000000 -> 1 0000000 -> negative zero? Because of this, we…