Programming Articles - Page 1499 of 3366

The diameter of a binary tree is the (left_height + right_height + 1) for each node. So in this method we will calculate (left_height + right_height + 1) for each node and update the result . The time complexity here stays O(n).Let us first define the struct that would represent a tree node that contains the data 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 newNode(int data) function that ... Read More

A matrix is said to be diagonally dominant matrix if for every matrix row, the diagonal entry magnitude of the row is larger than or equal to the sum of the magnitudes of every other non-diagonal entry in that row.Let us first define a constant int variable N with value 3 which represents our matrix dimensions.const int N = 3;The isDDM(int mat[N][N], int n) is a Boolean function that takes a copy of our matrix and the size of our matrix. Inside we iterate the rows and columns of our matrix using nested for loop. We then find the sum ... Read More

To consider the nodes that are passing between lines of slope -1. The diagonal traversal of the binary tree will be to traverse and print all those nodes present between these lines.Let us first define the struct that would represent a tree node that contains the data 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 data) function that takes an int value and assign it to the data ... Read More

To consider the nodes that are passing between lines of slope -1. The diagonal sum of the binary tree will be calculated by the sum of all nodes data that are present between these lines of reference.Let us first define the struct that would represent a tree node that contains the data 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 data) function that takes an int value and ... Read More

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