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Programming Articles
Page 2416 of 2547
Regular Expression "$" (dollar) Metacharacter in Java
The subexpression/metacharacter “$” matches the end of a line.Example 1import java.util.regex.Matcher; import java.util.regex.Pattern; public class EndWith { public static void main( String args[] ) { String regex = "Tutorialspoint$"; String input = "Hi how are you welcome to Tutorialspoint"; Pattern p = Pattern.compile(regex); Matcher m = p.matcher(input); int count = 0; while(m.find()) { count++; System.out.println("Number of matches: "+count); } } }OutputNumber of matches: 1Example 2The following Java program accepts ...
Read MoreRegular Expression "^" (caret) Metacharacter in Java
The subexpression/metacharacter “^” matches the beginning of a line. If you use this in a regular expression, it matches the sentence succeeding it in the input string.Example 1import java.util.regex.Matcher; import java.util.regex.Pattern; public class RegexExample { public static void main( String args[] ) { String regex = "^Hi how are you"; String input = "Hi how are you welcome to Tutorialspoint"; Pattern p = Pattern.compile(regex); Matcher m = p.matcher(input); int count = 0; while(m.find()) { count++; ...
Read MoreC++ Balanced expressions such that given positions have opening brackets
A balanced expression of parentheses is an expression that contains pairs of all sort of parentheses together in a correct order. this means that for every opening parentheses there is a closing parentheses in proper order of parentheses i.e. { }.Expression − {([][]{})({}[]{})}Output − balancedNow, in this problem we have to create all possiblebalanced expressions from the given number of brackets.And the condition is that the given position have opening brackets.In this problem, we are given an integer n and an array of position of the brackets of length 2n and we have to find the number of balanced expressions ...
Read MoreFind the smallest number X such that X! contains at least Y trailing zeros in C++
We have to take a number Y, we will find smallest number X, such that X! contains at least Y number of training zeros. For example, if Y = 2, then the value of X = 10. As X! = 3228800. It has Y number of zeros.We can solve this using binary search. The number of trailing zeros in N! is given by the count of the factors 5 in N!. X can be found using binary search in range [0, 5*Y]Example #include using namespace std; int factorCount(int n, int X) { if (X < n) ...
Read MoreFind the smallest and second smallest elements in an array in C++
Suppose we have an array of n elements. We have to find the first, second smallest elements in the array. First smallest is the minimum of the array, second smallest is minimum but larger than the first smallest number.Scan through each element, then check the element, and relate the condition for first, and second smallest elements conditions to solve this problem.Example #include using namespace std; int getTwoSmallest(int arr[], int n) { int first = INT_MAX, sec = INT_MAX; for (int i = 0; i < n; i++) { if (arr[i] < first) { ...
Read MoreFind the product of last N nodes of the given Linked List in C++
Consider we have few elements in a linked list. We have to find the multiplication result of last n number of elements. The value of n is also given. So if the list is like [5, 7, 3, 5, 6, 9], and n = 3, then result will be 5 * 6 * 9 = 270.The process is straight forward. We simply read the current element starting from left side, then add the elements into stack. After filling up the stack, remove n elements and multiply them with the prod. (initially prod is 1), when n number of elements are ...
Read MoreFind the product of first k nodes of the given Linked List in C++
Consider we have few elements in a linked list. We have to find the multiplication result of first k number of elements. The value of k is also given. So if the list is like [5, 7, 3, 5, 6, 9], and k = 3, then result will be 5 * 7 * 3 = 105.The processes is straight forward. We simply read the current element starting from left side, then multiply it with the prod. (initially prod is 1), when k number of elements are traversed, then stop.Example#include #include using namespace std; class Node{ public: ...
Read MoreFind the perimeter of a cylinder in C++
Suppose we have the diameter and the height of the cylinder, we have to find the perimeter of the cylinder. As the perimeter is the outline of two dimensional object, then we cannot find the perimeter of one three dimensional object directly. We can make a cross section of the cylinder, and convert it as rectangle, then find the perimeter. The two sides of the rectangular cross section are the diameter, and the height. So perimeter is −p=(2*d)+(2*h)Example#include using namespace std; int getCylinderPerimeter(int d, int h) { return (2*d) + (2*h); } int main() { int diameter = ...
Read MoreFind the Number which contain the digit d in C++
Consider we have a digit d, and the upper limit n. we have to find all numbers that contains d in range 0 to n. So if n = 20, and digit is 3, then the numbers will be [3, 13].To solve this problem, we will take every number as string, then if the digit is present in the string, the number will be printed, otherwise ignored.Example#include using namespace std; int getAllNumWithDigit(int n, int d) { string str = ""; str += to_string(d); char ch = str[0]; string p = ""; p += ch; for (int i = 0; i
Read MoreFind the Next perfect square greater than a given number in C++
Suppose we have a number n. our task is to find next perfect square number of n. So if the number n = 1000, then the next perfect square number is 1024 = 322.To solve this, we have get the square root of the given number n, then take the floor of it, after that display the square of the (floor value + 1)Example#include #include using namespace std; int justGreaterPerfectSq(int n) { int sq_root = sqrt(n); return (sq_root + 1)*(sq_root + 1); } int main() { int n = 1000; cout
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