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Programming Articles
Page 1411 of 2547
Primality Test in C++
In this problem, we are given a number N and our task is to check whether it is a prime number or not.Primality test s the algorithm that is used to check whether the given number is prime or not.Prime number is a number which can be divided by itself only. Example : 2, 3, 5, 7.Let’s take an example to understand our problem, Input: 11 Output: YesThere are multiple methods to check for primality test of a number.One simple method to check for primality is by checking the division of the number by all numbers less than N. If ...
Read MoreSum Root to Leaf Numbers in Python
Suppose we have a binary tree containing digits from 0-9 only, here all root-to-leaf path could represent a number.So if the tree is like −This is representing two paths 21 and 23, so the output will be 21 + 23 = 44.To solve this, we will follow these steps −Create one recursive function called dfs(), this will take root, and num. initially num = 0if the node is not nullnum := num * 10 + value of nodeif node right is not null and node left is not null, then’sum := sum + numnum := num / 10return from the ...
Read MorePrevious smaller integer having one less number of set bits in C++
In this problem, we are given an integer n. Our task is to print the largest number less than n which can be formed by changing one set bit of the binary representation of the number.Let’s take an example to understand the problemInput: n = 3 Output: 2 Explanation: (3)10 = (011)2 Flipping one set bit gives 001 and 010. 010 is greater i.e. 2.To solve this problem, we will have to flip the rightmost set bit and make it zero which will create the number the greatest possible number less than n that is found by flipping one bit ...
Read MorePrevious number same as 1's complement in C++
In this problem, we are given an integer n. Our task is to check weather the preceding number is equal to 1’s complement of the number.Let’s take a few examples to understand our problemInput: 12 Output: No Explanation: (12)10 = (1100)2 Preceding number 11 = (1011)2 1’s complement of 12 = (0011)2 Input: 4 Output: Yes Explanation: 4 = (100)2 Preceding number 3 = (011)2 1’s complement of 12 = (011)2To solve this problem, we can use a simple approach which is by comparing the previous number and the 1’s complement of the number.This approach is simple but consumes space ...
Read MoreCoin Change in Python
Suppose we have coins of different denominations and a total amount of money amount. We have to define one function to compute the fewest number of coins that we need to make up that amount. When that amount of money cannot be accommodated by any combination of the coins, return -1. So if the input is [1, 2, 5], and the amount is 11, the output is 3. This is formed using 5 + 5 + 1 = 11.To solve this, we will follow these steps −if amount = 0, then return 0if minimum of coins array > amount, then ...
Read MorePrevious greater element in C++
In this problem, we are given an array. Our task is to return the greatest element that precedes the current element in the array otherwise print -1.Let’s take an example to understand the problemInput: {6, 2, 7, 1, 5, 3} Output: -1, 6, -1, 7, 7, 7To solve this problem, an easy and obvious solution will be using nested loops that will check greater element in the preceding part of the array.Program to show the implementation of our solutionExample#include using namespace std; void preceddingGreatestElement(int arr[], int n){ cout = 0; j--) { if (arr[i]
Read MoreBinomial Random Variables in C++
Random variables are those variables that are an outcome of the outcomes of a process that has the probability of giving rise to multiple outcomes. For example, The variable denoting head or tail as an outcome on tossing a coin is a random variable.A binomial random variable is a special type of random variable whose value is related to an event that has a fixed probability of an outcome in an event.There are certain properties that are possessed by a binomial random variable that makes it special. These are a must for a variable to become a binomial random variable ...
Read MoreCharacter class: intersection - Java regular expressions
The character classes in Java regular expression is defined using the square brackets "[ ]", this subexpression matches a single character from the specified or, set of possible characters. For example the regular expression [abc] matches a single character a or, b or, c.The intersection variant of the character class allows you to match a character which is common in the ranges that have intersection relation between them.An intersection relation between ranges is defined using && i.e. the expression [a-z&&[r-u]] matches a single character from r to u.Exampleimport java.util.Scanner; import java.util.regex.Matcher; import java.util.regex.Pattern; public class RegexExample1 { public static ...
Read MoreClimbing Stairs in C++
There are n stairs. One person will go to 1st to nth stairs. Maximum how many stairs he/she can cross in one step is also given. With this information, we have to find possible ways to go to the nth stairs. Let us consider one can cross a maximum two stairs in each step. So we can find recursive relations to solve this problem. One can move to nth stair, either from (n-1)th stair or from (n-2)th stair. So ways(n) = ways(n-1) + ways(n-2).Suppose the number of stairs, say 10, the maximum number of stairs that can be jumped in ...
Read MoreCharacter class: subtraction - Java regular expressions
You can subtract one range from other and use it as new range. You can achieve this by using two variants of character classes i.e. negation and intersection.For example the intersection of ranges [a-l] and [^e-h] gives you the characters a to l as rage subtracting the characters [e-h]Exampleimport java.util.Scanner; import java.util.regex.Matcher; import java.util.regex.Pattern; public class RegexExample1 { public static void main(String[] args) { Scanner sc = new Scanner(System.in); System.out.println("Enter input text: "); String input = sc.nextLine(); String regex = "[a-l&&[^e-h]]"; //Creating a pattern ...
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