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Programming Articles
Page 2403 of 2547
Find the Nth term of the series where each term f[i] = f[i – 1] – f[i – 2] in C++
Suppose we have a series called f. Each term of f, follows this rule f[i] = f[i – 1] – f[i – 2], we have to find the Nth term of this sequence. f[0] = X and f[1] = Y. If X = 2 and Y = 3, and N = 3. The result will be -2.If we see this closely, there will be almost six terms before the sequence starts repeating itself. So we will find the first 6 terms of the series and then the Nth term will be the same as (N mod 6)th term.Example#include< iostream> using ...
Read MoreFind the altitude and area of an isosceles triangle in C++
Consider we have the side of the isosceles triangle, our task is to find the area of it and the altitude. In this type of triangle, two sides are equal. Suppose the sides of the triangle are 2, 2 and 3, then altitude is 1.32 and the area is 1.98. Altitude(h)=$$\sqrt{a^{2}-\frac{b^{2}}{2}}$$ Area(A)=$\frac{1}{2}*b*h$ Example #include #include using namespace std; float getAltitude(float a, float b) { return sqrt(pow(a, 2) - (pow(b, 2) / 4)); } float getArea(float b, float h) { return (1 * b * h) / 2; } int main() { float a = 2, b = 3; cout
Read MoreHow to detect duplicate values in primitive Java array?
To detect the duplicate values in an array you need to compare each element of the array to all the remaining elements, in case of a match you got your duplicate element.One solution to do so you need to use two loops (nested) where the inner loop starts with i+1 (where i is the variable of outer loop) to avoid repetitions in comparison.Exampleimport java.util.Arrays; import java.util.Scanner; public class DetectDuplcate { public static void main(String args[]) { Scanner sc = new Scanner(System.in); System.out.println("Enter the size of the array that is to ...
Read MoreHow to create array of strings in Java?
In Java, you can create an array just like an object using the new keyword. The syntax of creating an array in Java using new keyword −type[] reference = new type[10];Where, type is the data type of the elements of the array.reference is the reference that holds the array.And, if you want to populate the array by assigning values to all the elements one by one using the index −reference [0] = value1; reference [1] = value2;You can declare an array of Strings using the new keyword as &mius;String[] str = new String[5]; And then, you can populate the string ...
Read MoreFind minimum difference between any two element in C++
Suppose we have an array of n elements called A. We have to find the minimum difference between any two elements in that array. Suppose the A = [30, 5, 20, 9], then the result will be 4. this is the minimum distance of elements 5 and 9.To solve this problem, we have to follow these steps −Sort the array in non-decreasing orderInitialize the difference as infiniteCompare all adjacent pairs in the sorted array and keep track of the minimum oneExample#include #include using namespace std; int getMinimumDifference(int a[], int n) { sort(a, a+n); int min_diff = INT_MAX; for (int i=0; i
Read MoreHow to add elements to the midpoint of an array in Java?
Apache commons provides a library named org.apache.commons.lang3 and, following is the maven dependency to add library to your project. org.apache.commons commons-lang3 3.0 This package provides a class named ArrayUtils. You can add an element at a particular position in an array using the add() method of this class.Exampleimport java.util.Arrays; import org.apache.commons.lang3.ArrayUtils; public class AddingElementToMidPoint { public static void main(String args[]) { int[] myArray = {23, 93, 30, 56, 92, 39}; int eleToAdd = 40; int position= myArray.length/2; int [] result = ArrayUtils.add(myArray, position, eleToAdd); System.out.println(Arrays.toString(result)); } }Output[23, 93, 30, 40, 56, 92, 39]
Read MoreFind three closest elements from given three sorted arrays in C++
Suppose we have three sorted arrays A, B and C, and three elements i, j and k from A, B and C respectively such that max(|A[i] – B[i]|, |B[j] – C[k]|, |C[k] – A[i]|) is minimized. So if A = [1, 4, 10], B = [2, 15, 20], and C = [10, 12], then output elements are 10, 15, 10, these three from A, B and C.Suppose the size of A, B and C are p, q and r respectively. Now follow these steps to solve this −i := 0, j := 0 and k := 0Now do the following ...
Read MoreWhat is the difference between non-static methods and abstract methods in Java?
Following are the notable differences between non-static methods and abstract methods.Non-static (normal) methodsAbstract methodsThese methods contain a body.Abstract methods don’t have body these are ended with a semicolonYou can use normal method directly.You cannot use abstract methods directly, to use them you need to inherit them and provide body to these methods and use them.Example:public void display() { System.out.println("Hi"); }Example:public void display();
Read MoreFind pair with maximum difference in any column of a Matrix in C++
Suppose we have one matrix or order NxN. We have to find a pair of elements which forms maximum difference from any column of the matrix. So if the matrix is like −123535967So output will be 8. As the pair is (1, 9) from column 0.The idea is simple, we have to simply find the difference between max and min elements of each column. Then return max difference.Example#include #define N 5 using namespace std; int maxVal(int x, int y){ return (x > y) ? x : y; } int minVal(int x, int y){ return (x > y) ? ...
Read MoreFind the missing number in Geometric Progression in C++
Suppose we have an array that represents elements of geometric progression in order. One element is missing. We have to find the missing element. So if arr = [1, 3, 27, 81], output is 9, as 9 is missing.Using binary search, we can solve this problem. We will go to the middle element, then check whether the ratio between middle and next to the middle is same as common ratio or not. If not, then missing element is present between indices mid and mid + 1. If the middle element is the n/2th element in the GP, then missing element ...
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