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Page 1387 of 2109
Find the last digit when factorial of A divides factorial of B in C++
If we have two integers A and B, and B >= A, we have to compute the last digit of the B! / A! When the value of A = 2 and B = 4, then result is 2, 2! = 2 and 4! = 24, so 24/2 = 12. the last digit is 2.As we know that the last digit of factorial will be in set {0, 1, 2, 4, 6}, then follow these steps to solve this problem −We will find the difference between A and Bif diff >=5, then answer is 0otherwise, iterate from (A + 1) ...
Read MoreFind smallest values of x and y such that ax – by = 0 in C++
Suppose we have two values a and b. We have to find x and y, such that ax – by = 0. So if a = 25 and b = 35, then x = 7 and y = 5.To solve this, we have to calculate the LCM of a and b. LCM of a and b will be the smallest value that can make both sides equal. The LCM can be found using GCD of numbers using this formula −LCM (a,b)=(a*b)/GCD(a,b)Example#include #include using namespace std; void getSmallestXY(int a, int b) { int lcm = (a * b) / __gcd(a, b); cout
Read MoreFind sum of a number and its maximum prime factor in C++
Suppose we have a positive number n, and we have to find the sum of N and its maximum prime factor. So when the number is 26, then maximum prime factor is 13, so sum will be 26 + 13 = 39.Approach is straight forward. Simply find the max prime factor, and calculate the sum and return.Example#include #include using namespace std; int maxPrimeFact(int n){ int num = n; int maxPrime = -1; while (n % 2 == 0) { maxPrime = 2; n /= 2; } for (int i = 3; i 2) maxPrime = n; return maxPrime; } int getRes(int n) { int sum = maxPrimeFact(n) + n; return sum; } int main() { int n = 26; cout
Read MoreFind the center of the circle using endpoints of diameter in C++
Suppose we have two endpoints of diameter of a circle. These are (x1, y1) and (x2, y2), we have to find the center of the circle. So if two points are (-9, 3) and (5, -7), then the center is at location (-2, -2).We know that the mid points of two points are −$$(x_{m},y_{m})=\left(\frac{(x_{1}+x_{2})}{2},\frac{(y_{1}+y_{2})}{2}\right)$$Example#include using namespace std; class point{ public: float x, y; point(float x, float y){ this->x = x; this->y = y; } void display(){ cout
Read MoreFind the closest and smaller tidy number in C++
Suppose we have a number n, we have to find the closest and smaller tidy number of n. So a number is called tidy number, if all of its digits are sorted in non-decreasing order. So if the number is 45000, then the nearest and smaller tidy number will be 44999.To solve this problem, we will traverse the number from end, when the tidy property is violated, then we reduce digit by 1, and make all subsequent digit as 9.Example#include using namespace std; string tidyNum(string number) { for (int i = number.length()-2; i >= 0; i--) { ...
Read MoreFind the count of numbers that can be formed using digits 3 and 4 only and having length at max N in C++
Given a number N. We have to find the count of such numbers that can be formed using digit 3 and 4. So if N = 6, then the numbers will be 3, 4, 33, 34, 43, 44.We can solve this problem if we look closely, for single digit number it has 2 numbers 3 and 4, for digit 2, it has 4 numbers 33, 34, 43, 44. So for m digit numbers, it will have 2m values.Example#include #include using namespace std; long long countNumbers(int n) { return (long long)(pow(2, n + 1)) - 2; } int main() { int n = 3; cout
Read MoreFind the count of Strictly decreasing Subarrays in C++
Suppose we have an array A. And we have to find the total number of strictly decreasing subarrays of length > 1. So if A = [100, 3, 1, 15]. So decreasing sequences are [100, 3], [100, 3, 1], [15] So output will be 3. as three subarrays are found.The idea is find subarray of len l and adds l(l – 1)/2 to result.Example#include using namespace std; int countSubarrays(int array[], int n) { int count = 0; int l = 1; for (int i = 0; i < n - 1; ++i) { if ...
Read MoreFind the count of substrings in alphabetic order in C++
Suppose we have a string of length n. It contains only uppercase letters. We have to find the number of substrings whose character is occurring in alphabetical order. Minimum size of the substring will be 2. So if the string is like: “REFJHLMNBV”, and substring count is 2, they are “EF” and “MN”.So to solve this, we will follow these steps −Check whether str[i] + 1 is same as the str[i+1], if so, then increase the result by 1, and iterate the string till next character which is out of alphabetic order, otherwise continue.Example#include using namespace std; int countSubstr(string main_str) ...
Read MoreFind the fractional (or n/k – th) node in linked list in C++
Suppose we have a singly linked list and the number k. We have to write a function to find the (n/k)th element, where n is the number of elements in the list. For decimals, we will choose the ceiling values. So if the list is like 1, 2, 3, 4, 5, 6, and k = 2, then output will be 3, as n = 6 and k = 2, then we will print n/k th node so 6/2 th node = 3rd node that is 3.To solve this we have to follow some steps like below −Take two pointers called ...
Read MoreFind the next identical calendar year in C++
Suppose we have an year Y. Find next identical calendar year to Y. So the calendar of 2017 is identical with 2023.A year X is identical to given previous year Y if it matches these two conditions.x starts with the same day as year, If y is leap year, then x also, if y is normal year, then x also normal year.The idea is to check all years one by one from next year. We will keep track of number of days moved ahead. If there are total 7 moved days, then current year begins with same day. We also ...
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