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Articles by Prateek Jangid
Page 3 of 17
C++ program to Reorder the Given String to Form a K-Concatenated String
We are given a string and an integer k, and we need to reorder the characters in the string so that it becomes the concatenation of k similar substring. If not possible, output the result as "Impossible”. string = "malaalam"; K = 2; res = solve(s, K); Example (Using Maps) Let's have a string "mottom" and K=2. The given string can be represented as the concatenation of 2 substrings like tomtom, motmot omtomt, etc. As in all 3, two substrings are joined together when k = 2. Using the string, we can determine how many times each character occurs. ...
Read MoreC++ program to Replace Duplicates with Greater than Previous Duplicate Value
A series of integers is given in this article. Let's say we have an array of four elements without counting the repetitive elements, [2, 2, 5, 5, 7, 8, 7], and we have to make the array distinctive. Changing a value with one that is greater than the previous one is possible. In the above array, element 2 at index 1 becomes 3 as the next greater element. 5 at index 3 becomes 6 as the next greater and so on. So, in the end, our array becomes [2 3 5 6 7 8 9] and should minimize the sum ...
Read MoreC++ program to remove row or column wise duplicates from matrix of characters
We are given a 2D matrix with rows and columns. The matrix consists of elements in char data type. A method is devised to remove the elements which are duplicated in their respective rows or columns. In this method, we check if any element is repeating in its row or column for each character. If it is not repeated, we leave it as it was before. We can store the values occurring in each row and column in a map. After which, we can traverse again and take those values which are appearing only once in their row and column. ...
Read MoreFinding the second largest element in BST using C++
In a binary search tree (BST), the second largest element must be returned. In a binary tree, the second element is the largest element. According to the given BST, 13 is the second largest element. Now we are using the C++ approach to solve this problem. We can traverse the tree inorder, and by observation, we can observe that the second largest element in the given BST is 13. The inorder of the tree will be 1 3 4 6 7 8 10 13 14, and we can observe that the elements are in the sorted array. So we ...
Read MoreC++ program to find the shortest distance between two nodes in BST
In this article, we are given a BST (binary search tree), and we need to find the shortest distance between 2 given nodes in the BST. Let's have a tree and below are the following scenarios. Let’s assume some simple input and output scenarios Now we have to find the distances between nodes 4 and 13. int key1=13, key2=4; res = solve(root, min(key1, key2), max(key1, key2)); output = 6 The shortest distance is 6, which is through 4->6->3->8->10->14->13(the arrows show a path definition and not anything else). Let’s find another distance from two different nodes in the above ...
Read MoreC++ program to Replace a Node with Depth in a Binary Tree
Suppose we have a binary tree and we want to replace the depth of each node with its value. The depth of the node starts from 0 at the root node and increases by 1 for each level we go; for example, we have a binary tree like this; Here we replace, Node Value Depth 1 0 2 1 3 1 4 2 5 2 6 2 7 2 8 3 9 3 We do a simple ...
Read MoreReplace Each Node in Binary Tree With The Sum Of Its Inorder Predecessor And Successor Using C++
We are given a binary tree, and we need to replace all the elements with the sum of its inorder predecessor and successor. Inorder is a traversed path in a graph that reads in the order of left node – root node – right node. The method adds the elements in left and right nodes of the parent node and replaces the value with the obtained sum. Suppose we have a tree with the following formation and characters − We can find and store the inorder of the tree in an array. After that, we can again do an ...
Read MoreRencontres Number (Counting partial derangements) Using C++
Given two integers N and k, we need to count the number of derangements where k points are fixed at their position. Given constraints on k are between 0 and n as the number of fixed points when there are n points cannot be more than n. int N=4, k=2; res = solve(N, k); Note that at least conditions don’t hold on k. There has to be precisely and strictly k points on their original index. This is a mathematical problem. Not explaining the proof and explanation of mathematics, we as computer science majors can use the results ...
Read MoreRemove Edges Connected to a Node Such That The Three Given Nodes are in Different Trees Using C++
Suppose we are given a binary tree and three nodes in that binary. We have to disconnect one node entirely from the tree. Disconnecting that node leaves us with three different trees. Each of the three given nodes lies in one of them, or each of the given three nodes must not exist in the same tree. Disconnecting a node means that we will remove all the edges from this node to all other nodes. Example For example, let's say we have a tree with three nodes 18, 15, and 17 as shown below − If the task ...
Read MoreC++ program to represent the Fraction of Two Numbers in the String Format
We are given two integer numerators and a denominator. We need to represent the fraction of these two integers in string format. If a certain decimal is repeating, we need a bracket to show its repeating sequence. Algorithm (Steps) Following are the Algorithm/steps to be followed to perform the desired task − Determine the integral quotient (absolute part before to the decimal point) before determining the fractional portion. Insert the remainder (numerator % denominator) in a map with the key being the remainder and the value being the index position at which this remainder occurs to see if ...
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