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The "butterfly options strategy", also called "butterfly spreads" contain both bullish and bearish options. The trader in butterfly spreads has four options having the same expiry dates but three different strike prices. The trader buys two options contracts that have a higher and lower price, and two contracts with a price in between. The difference between high and low strike price is equal to the strike price in between.A butterfly strategy contains the following −Buying or selling of Call/Put optionsCombining four option contractsSame underlying assetSame expiry dateDifferent strike prices, with two contracts at same strike priceExplanationButterfly spreads work the best ... Read More
The prices of shares keep moving up and down in the market and hence, they are called "volatile" which means not constant over time. Implied volatility means how much the price of an option will move in a given period of time. The term is more applicable in the case of options contracts.Predicting volatility is a very important issue in finance. Investors often want to know how much an option or a stock can move up or down to reap the benefits and implied volatility can help them ascertain the basics of price movements.Implied Volatility – DefinitionImplies Volatility (or IV) ... Read More
Businesses need to take various investment decisions from time to time to stay functional and competitive. These decisions not only increase the competitive edge of a company but also add new dimensions to the existing manufacturing and financial position of a company.The investments needed for growth can be divided into the following four categories −Expansion InvestmentsDiversification InvestmentsModernization InvestmentsReplacement InvestmentsExpansion InvestmentsExpansion investment is made to increase the production of a certain product. It requires the firms to increase their capacity to manufacture or build a new production line to expand the current business volume. Expansion is often required when the demand ... Read More
Given an array, we have to find the number of pairs whose Bitwise OR is an odd number. Let's see the example.Inputarr = [1, 2]Output1 There is only one pair whose Bitwise OR is an odd number. And the pair is (1, 2).AlgorithmInitialise the array with random numbers.Initialise the count to 0.Write two loops to get the pairs of the array.Compute the bitwise OR between every pair.Increment the count if the result is an odd number.Return the count.ImplementationFollowing is the implementation of the above algorithm in C++#include using namespace std; int getOddPairsCount(int arr[], int n) { int count ... Read More
Given an array, we have to find the number of pairs whose sum is a power of 2. Let's see the example.Inputarr = [1, 2, 3]Output1 There is only one pair whose sum is a power of 2. And the pair is (1, 3).AlgorithmInitialise array with random numbers.Initialise the count to 0.Write two loops to get all the pairs of the array.Compute the sum of every pair.Check whether the sum is a power of 2 or not using bitwise AND.Increment the count if the count is a power of 2.Return the count.ImplementationFollowing is the implementation of the above algorithm in ... Read More
Given numbers N and K, we have to count the number of pairs whose sum is divisible by K. Let's see an example.InputN = 3 K = 2Output1 There is only one pair whose sum is divisible by K. And the pair is (1, 3).AlgorithmInitialise the N and K.Generate the natural numbers till N and store them in an array.Initialise the count to 0.Write two loops to get all pairs from the array.Compute the sum of every pair.If the pair sum is divisible by K, then increment the count.Return the count.ImplementationFollowing is the implementation of the above algorithm in C++#include ... Read More
The line equation that should be satisfied is y = mx + c. Given an array, m, and c, we have to find the number of order points satisfying the line equation. Let's see an example.Inputarr = [1, 2, 3] m = 1 c = 1Output2The pairs satisfying the line equation are2 1 3 2AlgorithmInitialise the array, m, and c.Write two loops to get all pairs from the array.Check whether the pair is satisfying the line equation or not.We can check the whether the equation is satisfied or not by substituting the values into the line equation.If the pair satisfies ... Read More
In this tutorial, we are going to write a program that finds the number non-negative integral solution of sum equation.The sum equation is x + y + z = n. You are given the number n, you need to find the number of solutions for the equation. Let's see an example.Input2Output6Solutions are0 0 2 0 1 1 0 2 0 1 0 1 1 1 0 2 0 0AlgorithmInitialise the number m.Initialise the count to 0.Write three nested loops to get all the combinations of three numbers.Check the validation of the equation.If the current numbers satisfies the equation, then increment ... Read More
Given an n-ary tree and a number, we have to count the number of nodes greater than the given number. Let's see an example.Inputtree = [[4], [1, 2], [3, 5]] n = 2Output3There are 3 nodes with values that are greater than n.AlgorithmInitialise the n-ary tree.Initialise the count to 0.Increment the count by 1 when you find a node whose value is greater than n.Get the children of the current node.Iterate over all the children and recursively call the same function to count nodes.Return the count.ImplementationFollowing is the implementation of the above algorithm in C++#include using namespace std; struct ... Read More
You are given an array and the index of the target element. We have to count the number of elements greater than the given element to its right. Let's see an example.Inputarr = [2, 3, 5, 1, 4, 2, 6] index = 3Output3The target index element is 1. There are three elements i.e..., 4, 2, 6 that are greater than 1 on its right side.AlgorithmInitialise the array and index of target element.If the index is greater than or equal to the length of the array, then return -1.Write a loop that iterates from the next element of the given index.Increment ... Read More