C++17 has extended existing if statement’s syntax. Now it is possible to provide initial condition within if statement itself. This new syntax is called "if statement with initializer". This enhancement simplifies common code patterns and helps users keep scopes tight. Which in turn avoids variable leaking outside the scope.ExampleLet us suppose we want to check whether given number is even or odd. Before C++17 our code used to look like this − Live Demo#include #include using namespace std; int main() {    srand(time(NULL));    int random_num = rand();    if (random_num % 2 == 0) {       cout

To check connectivity of a graph, we will try to traverse all nodes using any traversal algorithm. After completing the traversal, if there is any node, which is not visited, then the graph is not connected.For the directed graph, we will start traversing from all nodes to check connectivity. Sometimes one edge can have only outward edge but no inward edge, so that node will be unvisited from any other starting node.In this case the traversal algorithm is recursive DFS traversal.Input − Adjacency matrix of a graph0100000100000111000001000Output − The Graph is connected.Algorithmtraverse(u, visited) Input: The start node u and the ... Read More

Here we will see another interesting problem. We have to write a code that accepts a string, which has following criteria.Every group of consecutive 1s, must be of length 2every group of consecutive 1s must appear after 1 or more 0sSuppose there is a string like 0110, this is valid string, whether 001110, 010 are not validHere the approach is simple. we have to find the occurrences of 1, and check whether it is a part of sub-string 011 or not. If condition fails, for any substring then return false, otherwise true.Example Live Demo#include using namespace std; bool isValidStr(string str) ... Read More

Here we will see one interesting problem. We have to check whether a string has 0 in between a 1s or not. If not, then the string is valid, otherwise invalid.Suppose there are three strings −100011110100000111110001111101111From these three strings, only B is valid, because there is no 0 inside the stream of 1sTo solve this problem, we will find the index of first 1 present in the string, and also find the index of the last 1. Then we will check, is there any 0 from these two indices, if so, then return false, otherwise true (as valid)Example Live Demo#include ... Read More

The only safe way is to check for overflow before it occurs. There are some hacky ways of checking for integer overflow though. So if you're aiming for detecting overflow in unsigned int addition, you can check if the result is actually lesser than either values added. So for example, Exampleunsigned int x, y; unsigned int value = x + y; bool overflow = value < x; // Alternatively "value < y" should also workThis is because if x and y are both unsigned ints, if added and they overflow, their values can't be greater than either of them as ... Read More

Fleury’s Algorithm is used to display the Euler path or Euler circuit from a given graph. In this algorithm, starting from one edge, it tries to move other adjacent vertices by removing the previous vertices. Using this trick, the graph becomes simpler in each step to find the Euler path or circuit.We have to check some rules to get the path or circuit −The graph must be a Euler Graph.When there are two edges, one is bridge, another one is non-bridge, we have to choose non-bridge at first.sChoosing of starting vertex is also tricky, we cannot use any vertex as ... Read More

In this section, we will see how to find the 1’s complete of an integer. We can use the complement operator to do this task very fast, but it will make 32bit complemented value (4-bype integer). Here we want complement of n bit numbers.Suppose we have a number say 22. The binary equivalent is 10110. The complemented value is 01001 which is same as 9. Now the question comes, how to find this value? At first we have to find number of bits of the given number. Suppose the count is c (here c = 5 for 22). We have ... Read More

Here we will see another problem, where one string and another integer value say k is given. We have to find the number of substrings of length k, whose sum of ASCII values of characters is divisible by k.Suppose a string is “BCGABC”. And the value of k is 3. Here string BCG has ASCII sum 300, ABC has ASCII sum 294, both are divisible by k = 3.The approach is simple. At first we have to find the ASCII value of characters of first substring, whose length is k. We have to use the sliding window technique and subtract ... Read More

A number is given. We have to find two pairs, that can represent the number as sum of two cubes. So we have to find two pairs (a, b) and (c, d) such that the given number n can be expressed as n = a3 + b3 = c3 + d3The idea is simple. Here every number a, b, c and d are all less than n1/3. For every distinct pair (x, y) formed by number less than n1/3, if their sum (x3 + y3) is equal to the given number, we store them into hash table with the sum ... Read More

In this section we will see how to get number of pairs using the given GCD and LCM values. Suppose the GCD and LCM values are 2 and 12. Now the possible pairs of numbers are (2, 12), (4, 6), (6, 4) and (12, 2). So our program will find the count of pairs. That is 4.Let us see the algorithm to understand what will be the technique to solve this problem.AlgorithmcountPairs(gcd, lcm): Begin    if lcm is nit divisible by gcd, then       return 0    temp := lcm/gcd    c := primeFactorCount(temp)    res := shift ... Read More