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C++ Articles
Page 234 of 597
Construct a frequency array of digits of the values obtained from x^1, x^2, ....., x^nn in C++
Suppose we have two integers x and n. We have to find the array such that it contains the frequency of index numbers occurring in (x^1, x^2, … x^(n – 1), x^n). So if x = 15 and n = 3, then output will be [0, 1, 2, 2, 0, 3, 0, 1, 0, 0]. As we know that x^1 to x^n, the values are 15, 225 and 3375. So the frequency array is 0, 1, 2, 2, 0, 3, 0, 1, 0, 0To solve this, we will follow these steps −Maintain the frequency count array to store the counts ...
Read MoreConstruct a graph from given degrees of all vertices in C++
Suppose we have a list of vertices, and their degrees are given. We have to generate one undirected graph from that degree sequence. It will not include loop or multiple edges. So if the degree sequence is like [2, 2, 1, 1], then the graph can be likeTo solve this, we will follow these steps −Define adjacency matrix adj to store the graphfor each vertex i, dofor each vertex j that is valid, and next to iif the degree of vertex i and j are more than zero, then connect themdisplay the matrix.Example#include #include using namespace std; void ...
Read MoreConstruct BST from given preorder traversal - Set 2 in C++
Suppose we have one pre order traversal. From this traversal. we have to generate the tree So if the traversal is like [10, 5, 1, 7, 40, 50], then the tree will be like −To solve this, we will follow these steps −Create empty stackMake the first value as root, and push it into stack.Now keep pooping while the stack is not empty and the next value is greater than stack top element, make this as right child of the last popped node. Now push the new node to the stack.When the next value is less than the top element, ...
Read MoreConstruct BST from its given level order traversal in C++
Suppose we have one level order traversal. From this traversal. we have to generate the tree So if the traversal is like [7, 4, 12, 3, 6, 8, 1, 5, 10], then the tree will be like −To solve this, we will use recursive approach. The first element will be the root. The second element will be the left child, and third element will be right child (If the condition of BST satisfies), this property will be satisfied for all elements. So we will follow these steps −At first we have to take the first element of the array, and ...
Read MoreConvert a given Binary tree to a tree that holds Logical AND property on C++
In this tutorial, we will be discussing a program to convert a given Binary tree to a tree that holds Logical AND property.For this we will be provided with a binary tree. Our task is to convert it into a tree that holds the logical AND property means that a node has a value of the AND operation of its children nodes. Note that every node can have a value either zero or one.Example#include using namespace std; //node structure of binary tree struct Node{ int data; struct Node* left; struct Node* right; }; //creation of a new ...
Read MoreConvert a given Binary Tree to Doubly Linked List (Set 1) in C++
In this tutorial, we will be discussing a program to convert a binary tree to a doubly linked list.For this we will be provided with a binary tree. Our task is to convert it into a doubly linked list such that the left and right pointers become the previous and next pointers. Also the sequential order of the doubly linked list must be equal to the inorder traversal of the binary tree.For this we are having a very straight forward approach. We will be traversing the binary tree in in order way making the nodes of the doubly linked list ...
Read MoreConvert a given Binary Tree to Doubly Linked List (Set 2) in C++
In this tutorial, we will be discussing a program to convert a binary tree to a doubly linked list.For this we will be provided with a binary tree. Our task is to convert it into a doubly linked list such that the left and right pointers become the previous and next pointers. Also the sequential order of the doubly linked list must be equal to the inorder traversal of the binary tree.For this we are having a different approach. We will be traversing the binary tree in reverse inorder way. Along with we will be creating new nodes and moving ...
Read MoreConvert a normal BST to Balanced BST in C++
In this tutorial, we will be discussing a program to convert a normal binary search tree to balanced binary search tree.For this we will be provided with a skewed binary search tree either left or right. Our task is to convert it into a balanced binary search tree following a certain set of rules.Example#include using namespace std; //node structure of tree struct Node{ int data; Node* left, *right; }; //traversing tree and storing node pointers //in vector nodes void store_nodes(Node* root, vector &nodes){ if (root==NULL) return; store_nodes(root->left, nodes); nodes.push_back(root); store_nodes(root->right, ...
Read MoreConvert a number into negative base representation in C++
In this tutorial, we will be discussing a program to convert a number into its negative base representation.For this we will be provided with a number and the corresponding negative base. Our task is to convert the given number into its negative base equivalent. We are allowing only values between -2 and -10 for negative base values.Example#include using namespace std; //converting integer into string string convert_str(int n){ string str; stringstream ss; ss > str; return str; } //converting n to negative base string convert_nb(int n, int negBase){ //negative base equivalent for zero is zero ...
Read MoreConvert a number m to n using minimum number of given operations in C++
In this tutorial, we will be discussing a program to convert a number m to n using minimum number of given operations.For this we will be provided with two integers m and n. Our task is to convert the integer m to n using the given operations least times.Allowed operations −Multiply the given number by 2Subtract one from the given numberExample#include using namespace std; //finding minimum number of operations required int convert(int m, int n){ if (m == n) return 0; if (m > n) return m - n; //can't convert in ...
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