C++ Articles - Page 511 of 719

Here we will see how to find the nth term in Stern’s Diatomic series. The series is like 0, 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, … This is also known as fusc function. This series can be defined as −𝑝(𝑛)=$p\lgroup\frac{n}{2}\rgroup$ 𝑤ℎ𝑒𝑛 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛𝑝(𝑛)=$p\lgroup\frac{n-1}{2}\rgroup+p\lgroup\frac{n+1}{2}\rgroup$ 𝑤ℎ𝑒𝑛 𝑛 𝑖𝑠 𝑜𝑑𝑑𝑝(0)=0 𝑎𝑛𝑑 𝑝(1)=1Here we will use the Dynamic programming approach to reduce the number of computations. After saving the base case for p(0) and p(1), we will iterate from index i = 2 to n, and compute p(i)Example Live Demo#include using namespace std; int findTerm(int ... Read More

Suppose we have two integers N and D. We have to find a set of N integers, where the difference between their sum and product is the same as D. Suppose the N = 3, and D = 5, then the output will be 1, 2, 8. Here the sum is 1 + 2 + 8 = 11, and product is 1 * 2 * 8 = 16, the difference between 16 and 11 is 5.We have to solve this problem; we will use one tricky method. Here we will try to find N–2 number of 1s, one 2, and ... Read More

In this tutorial, we will be discussing a program to print out 2D shapes.For this we will be provided with the various parameters required to make a shape such as radius, side length and side breadth, etc. And our task is to print a shape accordingly with no thickness.Example Live Demo#include using namespace std; void print_circle(int radius){    for (int i = 0; i

Suppose we have three values, a, b and x. We have to find one multiple of x, that is nearest to ab. Suppose the numbers are x = 4, a = 3, b = 3, then the output will be 28, as this is nearest to 33 = 27The approach is simple; we have to follow these conditions −If b < 0, and a = 1, then ab turns out to be 1 and hence, the closest multiple of x becomes either 0 or x.If b < 0 and a > 1, then, ab, turns out to be less than ... Read More

Suppose we have four integers a, b, c and k. We have to find the minimum positive value x, such that the following equation satisfies −𝑎𝑥2+𝑏𝑥+𝑐 ≥𝑘If a = 3, b = 4, c = 5 and k = 6, then output will be 1To solve this, we will use the bisection approach. The lower limit will be 0 since x has to be a minimum positive integer.Example Live Demo#include using namespace std; int getMinX(int a, int b, int c, int k) {    int x = INT8_MAX;    if (k

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

In this tutorial, we will be discussing a program to print ‘N’ alphabet using the number pattern from 1 to n.For this we will have to print the english alphabet N. Our task is to determine the size of the letter and print it back using the numbers from 1 to n.Example Live Demo#include using namespace std; //printing the letter N void print_N(int N){    int index, side_index, size;    int Right = 1, Left = 1, Diagonal = 2;    for (index = 0; index < N; index++) {       cout

Suppose, we have one integer N > 0. The task is to find the maximum product of digits among numbers less than or equal to N. If the N is 390, then the result is 216, as the number 389 is making maximum product 3 * 8 * 9 = 216.To solve this problem, we will use the recursive approach. So if N = 0, then return 1, if the number N < 10, then return N, otherwise return max(max_product(N/10) * (N mod 10), max_product((N/10) - 1)*9)Example Live Demo#include using namespace std; int max_product(int N) {    if (N == 0) ... Read More

In this tutorial, we will be discussing a program to multiply two matrices.For this we will be given with two matrices and our task is to print the product of two those matrices. The only condition is that the number of columns of first matrix should be equal to the number of rows of the second matrix.Example Live Demo#include using namespace std; #define N 4 //multiplying the elements of both matrices void calc_product(int mat1[][N], int mat2[][N], int res[][N]){    int i, j, k;    for (i = 0; i < N; i++) {       for (j = 0; ... Read More

Suppose, one binary tree is given. It has positive and negative nodes. We have to find the maximum product at each level of it.Consider this is the tree, so the product of level 0 is 4, product of level 1 is 2 * -5 = -10, and product of level 2 is -1 * 3 * -2 * 6 = 36. So this is the maximum one.To solve this, we will perform the level order traversal of the tree, during traversal, process doing the nodes of different levels separately. Then get the maximum product.Example Live Demo#include #include using namespace std; class ... Read More