Article Categories
- All Categories
-
Data Structure
-
Networking
-
RDBMS
-
Operating System
-
Java
-
MS Excel
-
iOS
-
HTML
-
CSS
-
Android
-
Python
-
C Programming
-
C++
-
C#
-
MongoDB
-
MySQL
-
Javascript
-
PHP
-
Economics & Finance
Articles on Trending Technologies
Technical articles with clear explanations and examples
Camelcase Matching in C++
Suppose we have a list of queries, and a pattern, we have to return an answer that will be list of booleans, where answer[i] is true if and only if queries[i] matches the pattern. A query word matches a given pattern when we can insert lowercase letters to the pattern word so that it equals the query.So if the input is like ["FooBar", "FooBarTest", "FootBall", "FrameBuffer", "ForceFeedBack"] and pattern = "FB", then the results will be [true, false, true, false, false].To solve this, we will follow these steps −Define a method called insertNode(), this takes head and string scurr := ...
Read MoreCheck If Word Is Valid After Substitutions in C++
Suppose we are given that the string "abc" is valid. So from any valid string V, we can divide V into two pieces X and Y such that X + Y is same as V. (X or Y may be empty.). Then, X + "abc" + Y is also valid. So for example S = "abc", then examples of valid strings are: "abc", "aabcbc", "abcabc", "abcabcababcc". And some examples of invalid strings are: "abccba", "ab", "cababc", "bac". We have to check true if and only if the given string S is valid.So if the input is like “abcabcababcc”, then that ...
Read MoreAs Far from Land as Possible in C++
Suppose we have one N x N grid containing only values like 0 and 1, where 0 represents water and 1 represents the land, we have to find a water cell such that its distance to the nearest land cell is maximized and return the distance. Here we will use the Manhattan distance − the distance between two cells (x0, y0) and (x1, y1) is |x0 - x1| + |y0 - y1|. If no land or water is present in the grid, then return -1.101000101Then the output will be 2, as the cell (1, 1) is as far as possible ...
Read MoreXOR Cipher in C++
XOR cipher or XOR encryption is a data encryption method that cannot be cracked by brute-force method.Brute-force method is a method of random encryption key generation and matching them with the correct one.To implement this encryption method, we will define an encryption key(random character) and perform XOR of all characters of the string with the encryption key. This will encrypt all characters of the string.Program to show the implementation of encryption −Example#include #include using namespace std; void XORChiper(char orignalString[]) { char xorKey = 'T'; int len = strlen(orignalString); for (int i = 0; i < len; i++){ ...
Read MoreInvalid Transactions in C++
Suppose there are some transactions. A transaction is possibly invalid if −The amount exceeds $1000, or;If it occurs within (and including) 60 minutes of another transaction with the same name in a different city.Here each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction. We have a list of transactions, find a list of transactions that are possibly invalid. So if the input is like ["alice, 20, 800, mtv", "bob, 50, 1200, mtv"], then the answer will be ["bob, 50, 1200, mtv"].To solve this, we will follow these steps ...
Read MoreXOR counts of 0s and 1s in binary representation in C++
In this problem, we are given a number. Our task is to find the XOR of the count of 0s and 1s in the binary representation of the number.Let’s take an example to understand the problem, Inputn = 9Output0Explanationbinary = 1001 Count of 0s = 2 Count of 1s = 2 2 ^ 2 = 0To solve this problem, we will first convert the number of its binary equivalent and then iterating over each bit of the number, count 0s, and 1s and then find XOR of the count of 0s and count of 1s.Program to illustrate the above solution, ...
Read MoreMaximum Difference Between Node and Ancestor in C++
Suppose we have the root of a binary tree, we have to find the maximum value V for which there exists different nodes A and B where V = |value of A – value of B| and A is an ancestor of B. So if the tree is like −Then the output will be 7. The ancestor node differences are like [(8 - 3), (7 - 3), (8 - 1), (10-13)], among them the (8 - 1) = 7 is the maximum.To solve this, we will follow these steps −initially define ans 0define one method called solve(), this will take ...
Read MoreK-Concatenation Maximum Sum in C++
Suppose we have an integer array arr and one integer k, we have to change the array by repeating it k times. So if arr = [1, 2] and k = 3 then the modified array will be [1, 2, 1, 2, 1, 2].Now we have to find the maximum sub-array sum in the modified array. Note that the length of the sub-array can be 0 and its sum in that case is 0. As the answer may be very large, find the answer modulo 10^9 + 7.So if the input is like [1, -2, 1] and k = 5, ...
Read MoreReplace the Substring for Balanced String in C++
Suppose we have a string containing only 4 kinds of characters 'Q', 'W', 'E' and 'R'. A string will be balanced if each of its characters appears n/4 times where n is the length of the string. Find the minimum length of the substring that can be replaced with any other string of the same length to make the original string is balanced. So if s = “QQWE”, then output will be 1. This is because we can replace Q to R, so that “RQWE”, that is balanced.Return 0 if the string is already balanced.To solve this, we will follow ...
Read MoreHandling the Divide by Zero Exception in C++
In this tutorial, we will be discussing how to handle the divide by Zero exception in C++.Division by zero is an undefined entity in mathematics, and we need to handle it properly while programming so that it doesn’t return at error at the user end.Using the runtime_error classExample#include #include using namespace std; //handling divide by zero float Division(float num, float den){ if (den == 0) { throw runtime_error("Math error: Attempted to divide by Zero"); } return (num / den); } int main(){ float numerator, denominator, result; numerator = 12.5; denominator = 0; try { result = Division(numerator, denominator); cout
Read More