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To find the length for the diagonal of a regular pentagon we put the value of side in (1+√5)side/2 = (1+2.24)side/2.ExampleLet us see the following implementation to get the regular Heptagon diagonal from its side − Live Demo#include using namespace std; int main(){ float side = 5; if (side < 0) return -1; float diagonal = (1+2.24) *(side/2); cout
The regular hexagons are comprised of six equilateral triangles so the diagonal of a regular hexagon would be 2*side.ExampleLet us see the following implementation to get the regular Heptagon diagonal from its side − Live Demo#include using namespace std; int main(){ float side = 12; if (side < 0) return -1; float diagonal = 2*side; cout
To use Deterministic Finite Automaton(DFA) to find strings that aren’t ending with the substring “THE”. We should keep that in mind that any variation of the substring “THE” like “tHe”, “The” ,”ThE” etc should not be at the end of the string.First, we define our dfa variable and initialise it to 0 which keeps our track of state. It is incremented on each character matched.int dfa = 0;The begin(char c) method takes a character and checks if its ‘t’ or ‘T’ and go to first state i.e 1.void begin(char c){ if (c == 't' || c == 'T') ... Read More
The Deterministic Finite Automaton(DFA) is used for checking if a number is divisible by another number k or not. The algorithm is useful because it can also find the remainder if the number isn’t divisible.In DFA based division we build a DFA table with k states. We consider binary representation of the number so there is only 0 and 1 in each state in DFA.The createTransTable(int k, int transTable[][2]) function is used for creating the transTable and storing the states in it. It takes the number k by which the number is to be divisible and transTable[][2] which is an ... Read More
In a N sided polygon if two children are standing on A and B vertex then we need to determine the vertex number where another person should stand so that there should be minimum number of jumps required by that person to reach A and B both.Two conditions to note here are that the polygon vertices are numbered in a clockwise manner and we will always choose the least numbered vertex in case there are multiple answers.The vertexPosition(int sides, int vertexA, int vertexB) takes the no. of sides of the polygon and the position of vertex A and B. The ... Read More
The objective is to determine the number of squares a line will pass through given two endpoints (x1, y1) and (x2, y2).To find the number of squares through which our line pass we need to find : difference between the x points (dx) = x2-x1, difference between the y points (dy) = y2-y1, adding the dx and dy and subtracting by their gcd (result) = dx + dy – gcd(dx, dy).The unitSquares(int x1, int y1, int x2, int y2) function takes four values x1, y1 and x2, y2. The absolute difference between the x2 and x1 and the absolute difference ... Read More
The determinant of a matrix can be calculated only for a square matrix by multiplying the first row cofactor by the determinant of the corresponding cofactor and adding them with alternate signs to get the final result.$$A = \begin{bmatrix}a & b & c\d & e &f \g & h &i \ \end{bmatrix} |A| = a(ei-fh)-b(di-gf)+c(dh-eg)$$First we have the determinantOfMatrix(int mat[N][N], int dimension) function that takes the matrix and the dimension value of the matrix. If the matrix is of only 1 dimension then it returns the [0][0] matrix value. This condition is also used as a base condition since we ... Read More
A number whose sum of its digits powered with its respective position equals to the number itself is called a disarium number.The noOfDigits(int num) function takes the number and return the number of digits by constantly dividing the number by 10 while there is only ones place left. On each iteration the digits variable is incremented to keep the digits track and is returned once the while loop ends.int noOfDigits(int num){ int digits = 0; int temp = num; while (temp){ temp= temp/10; digits++; } return digits; }Next, isDisarium(int ... Read More
Let us first define the struct that would represent a tree node that contains the int key and its left and right node child. If this is the first node to be created then it’s a root node otherwise a child node.struct Node { int data; struct Node *leftChild, *rightChild; };Next we create our createNode(int key) function that takes an int key value and assign it to the key member of the node. The function returns the pointer to the created struct Node. Also, the left and right child of the newly created node are set to null.Node* ... Read More
Let us first define the struct that would represent a tree node that contains a character key and a vector of Node *.struct Node{ char key; vector children; };Next we create our createNode(int key) function that takes an int key value and assign it to the key member of the node. The function returns the pointer to the created struct node.Node *createNode(int key){ Node *node = new Node; node->key = key; return node; }Our depthOfTree(struct Node* root) function takes the root node as parameter. If the root is NULL then the depth is returned as ... Read More