Programming Articles - Page 2522 of 3363

Here we will see what are the differences of sizeof and the alignof operator in C++. The alognof() operator is introduced in C++11.The alignof() operator is used to get the alignment in bytes. It requires instances of type. the type is either complete type or a reference type. There is another operator called the sizeof() operator, that returns the size of one type. For normal datatypes the sizeof and the alignof returns the same value. For some user defined datatype, the alignof returns some different value. Let us see the example to get the idea.Example Live Demo#include using namespace std; struct ... Read More

Here we will see one problem, where we will add two n digit numbers but the carry will not be propagated. We can understand this concept through an example −So we can see that here only digits are getting added and answer is placed. Here is one trick. We have to scan the numbers from right to left. So the sum of 3+2 = 6 will be calculated first, but it will be placed at the end. So we will use stack to store intermediate results.AlgorithmnoPropagateCarry(a, b)begin    size = max of length of a and length of b   ... Read More

A JComboBox is a Swing component that has a built-in left-click menu. In this article, we will learn how to show a popup menu when the user right-clicks on a JComboBox in Java.  What is a JComboBox? A JComboBox is a subclass of the JComponent class that displays a drop-down list and gives users options that they can select only one item at a time. A JComboBox can be editable or read-only. The getSelectedItem() Method A getSelectedItem() method can be used to get the selected or entered item from a combo box. What is a Popup Menu? A popup menu is ... Read More

Here we will see how to calculate the remainder of array multiplication after dividing the result by n. The array and the value of n are supplied by the user. Suppose the array is like {12, 35, 69, 74, 165, 54} so the multiplication will be (12 * 35 * 69 * 74 * 165 * 54) = 19107673200. Now if we want to get the remainder after diving this by 47 it will be 14.As we can see this problem is very simple. we can easily multiply the elements then by using modulus operator, it can get the result. ... Read More

In this section we will see how we can get the product of unique prime factor of a number in an efficient way. There is a number say n = 1092, we have to get product of unique prime factors of this. The prime factors of 1092 are 2, 2, 3, 7, 13. So the unique prime factors are {2, 3, 7, 13}, the product is 546. To solve this problem, we have to follow this rule −When the number is divisible by 2, then multiply 2 with product, and divide the number by 2 repeatedly, then next 2s will ... Read More

Here we will see how the brick sort works. The Brick sort is one modification of bubble sort. This algorithm is divided into two parts. These parts are odd part and even parts. In the odd part we will use the bubble sort on odd indexed items, and in the even part we will use the bubble sort on even indexed elements. Let us see the algorithm to get the idea.AlgorithmbrickSort(arr, n)begin    flag := false    while the flag is not true, do       flag := true       for i := 1 to n-2, increase ... Read More

Here we will see how to find the vertex, focus directrix of a parabola using C or C++ program. To get these parameters we need the general equation of a parabola. The general formula is −𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐The values of a, b and c are given.The formula for the vertex −The formula for the focus −The formula for the Directrix - y −Example Live Demo#include using namespace std; void getParabolaDetails(float a, float b, float c) {    cout

Here we will see how to split an array, and add the first part after splitting at the end position. Suppose the array contents are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. We want to cut this intro two parts. The first part is from index 0 to 3 (split size 4), and second part is rest. After adding the first part at the end, the array elements will be like this {4, 5, 6, 7, 8, 9, 0, 1, 2, 3}. To solve this problem, we will follow this algorithm.AlgorithmsplitArray(arr, n, k)begin    for i := ... Read More

In this section we will see how we can get the sum of all even prime factors of a number in an efficient way. There is a number say n = 480, we have to get all factor of this. The prime factors of 480 are 2, 2, 2, 2, 2, 3, 5. The sum of all even factors is 2+2+2+2+2 = 10. To solve this problem, we have to follow this rule −When the number is divisible by 2, add them into the sum, and divide the number by 2 repeatedly.Now the number must be odd. So we will ... Read More

In this section we will see another approach of famous bubble sort technique. We have used bubble sort in iterative manner. But here we will see recursive approach of the bubble sort. The recursive bubble sort algorithm is look like this.AlgorithmbubbleRec(arr, n)begin    if n = 1, return    for i in range 1 to n-2, do       if arr[i] > arr[i+1], then          exchange arr[i] and arr[i+1]       end if    done    bubbleRec(arr, n-1) endExample Live Demo#include using namespace std; void recBubble(int arr[], int n){    if (n == 1)     ... Read More