Server Side Programming Articles - Page 1998 of 2650

Suppose, we have an integer N, we have to find the number of integers 1 < x < N, for which x and x + 1 has same number of positive divisors. So if N = 3, then output will be 1, as divisor of 1 is 1, divisor of 2 is 1 and 2, and divisor of 3 is 1 and 3.To solve this, we will find the number of divisors of all numbers below N, and store them in an array. Then count number of integers x such that x, such that x + 1 have the same ... Read More

Suppose, we have two integers x and n, our task is to search for the first consecutive stream of 1s (32-bit binary) which is greater than or equal to the value of n in length and return its position. If no such string exists, then return -1. For example, if x = 35, and n = 2, then result will be 31. The binary representation of 35 in a 32-bit integer is like −00000000000000000000000000100011. So two consecutive 1s are present at index 31, so the answer is 31.To solve this problem, we have to find the number of leading zeros, ... Read More

Suppose, we have a number n. Our task is to find the number of integers from 1 to n, which contains digits 0s and 1s only. So if n = 15, then output will be. As the numbers are 1, 10, 11To solve this, we will create integers using 0s and 1s using recursive function. Following code will help us to understand this better.Example Live Demo#include using namespace std; int numberOfValues(int p, int n) {    if (p > n)       return 0;    return 1 + numberOfValues(p * 10, n) + numberOfValues(p * 10 + 1, n); } int main() {    int n = 120;    cout

Suppose we have three arrays with some elements. We have to find all the common elements that are present in these three arrays. Suppose these elements are [10, 12, 15, 20, 25], [10, 12, 13, 15] and [10, 12, 15, 24, 25, 26], then the common elements in these three arrays are 10, 12 and 15.Suppose current element traversed in array A1 be x, A2 be y and A3 be z. We can have the following cases for them −If x, y, and z are same, then we will print any of them, and increase each array elements by 1When ... Read More

Suppose, we have an integer N, We have to find the number of all possible distinct binary strings of the length N, which have at least three consecutive 1s. So if n = 4, then the numbers will be 0111, 1110, 1111, so output will be 3.To solve this, we can use the Dynamic programming approach. So DP(i, x) indicates number of strings of length i with x consecutive 1s in position i + 1 to i + x. Then the recurrence relation will be like −DP(i, x) = DP(i – 1, 0) + DP(i – 1, x + 1).The ... Read More

Suppose we have three linked lists. We have to find all the common elements that are present in these three linked lists. Suppose these lists are [10, 12, 15, 20, 25], [10, 12, 13, 15] and [10, 12, 15, 24, 25, 26], then the common elements in these three lists are 10, 12 and 15.We will use the hashing technique to solve this problem. To solve this, we have to follow these steps −Create an empty hash table, and go through each element in the first table, and insert the elements, and mark the frequency as 1Iterate through the second ... Read More

Suppose we have two integers n and m. We have to find the number closest to n and divide by m. If there are more than one such number, then show the number which has maximum absolute value. If n is completely divisible by m, then return n. So if n = 13, m = 4, then output is 12.To solve this, we can follow this steps −let q := n/m, and n1 := m*qif n * m > 0, then n2 := m * (q + 1), otherwise n2 := m * (q - 1)if |n – n1| < ... Read More

Here we will see how to find the closest value for every element in an array. If an element x has the next element that is larger than it, and also present in the array, then that will be the greater value of that element. If the element is not present, then return -1. Suppose the array elements are [10, 5, 11, 6, 20, 12], then the greater elements are [11, 6, 12, 10, -1, 20]. As 20 has not greater value in the array, then print -1.To solve this, we will use the settings in C++ STL. The set ... Read More

Suppose we have two integers P and Q. We have to find smallest number K, such that K mod P = 0 and Q mod K = 0. Otherwise print -1. So if the P and Q are 2 and 8, then K will be 2. As 2 mod 2 = 0, and 8 mode 2 = 0.In order for K to be possible, Q must be divisible by P. So if P mod Q = 0 then print P otherwise print -1.Example Live Demo#include using namespace std; int getMinK(int p, int q) {    if (q % p == 0)    return p;    return -1; } int main() {    int p = 24, q = 48;    cout

Consider we have the selling price, and percentage of profit or loss is given. We have to find the cost price of the product. The formula is like below −$$Cost Price=\frac{Sell Price∗100}{100+percentage profit}$$ $$Cost Price=\frac{Sell Price∗100}{100+percentage loss}$$Example Live Demo#include using namespace std; float priceWhenProfit(int sellPrice, int profit) {    return (sellPrice * 100.0) / (100 + profit); } float priceWhenLoss(int sellPrice, int loss) {    return (sellPrice * 100.0) / (100 - loss); } int main() {    int SP, profit, loss;    SP = 1020;    profit = 20;    cout