Suppose we have three numbers a, b, c, we have to check whether a + b = c, after removing all 0s from the numbers or not. Suppose the numbers are a = 102, b = 130, c = 2005, then after removing 0s, the numbers will be a + b = c : (12 + 13 = 25) this is trueWe will remove all 0s from a number, then we will check after removing 0s, a + b = c or not.Example Live Demo#include #include using namespace std; int deleteZeros(int n) {    int res = 0;   ... Read More

Here we will see whether a number is sandwiched between primes or not. A number is said to be sandwiched between primes when the number just after it, and just below it is prime numbers. To solve this, check whether n-1 and n+1 are prime or not.Example Live Demo#include #include #define N 100005 using namespace std; bool isPrime(int n) {    if (n == 0 || n == 1)       return false;    for (int i=2;i

Here we will see one program, that can check whether a number is magic number or not. A number is said to be magic number, when the recursive sum of the digits is 1. Suppose a number is like 50311 = 5 + 0 + 3 + 1 + 1 = 10 = 1 + 0 = 1, this is magic number.To check whether a number is magic or not, we have to add the digits until a single-digit number is reached.Example Live Demo#include using namespace std; int isMagicNumber(int n) {    int digit_sum = 0;    while (n > ... Read More

Suppose we want to find the result after multiplying two numbers A and B. We have to check whether the multiplied value will exceed the 64-bit integer or not. If we multiply 100, and 200, it will not exceed, if we multiply 10000000000 and -10000000000, it will overflow.To check this, we have to follow some steps. These are like below −Steps −If anyone of the numbers is 0, then it will not exceedOtherwise, if the product of two divided by one equals to the other, then it will not exceedFor some other cases, it will exceed.Example Live Demo#include #include ... Read More

We have the sum and gcd of two numbers a and b. We have to find both numbers a and b. If that is not possible, return -1. Suppose the sum is 6 and gcd is 2, then the numbers are 4 and 2.The approach is like, as the GCD is given, then it is known that the numbers will be multiples of it. Now there following stepsIf we choose the first number as GCD, then the second one will be sum − GCDIf the sum of the numbers is chosen in the previous step is the same as the ... Read More

We have a string with three colors (G, B, Y). We have to find the resulting color based on these relations −B * G = YY * B = GG * Y = BSuppose the string is “GBYGB” is B. If the string is “BYB”, then it will be Y.The approach is simple; we will take the string. Compare each alphabet with adjacent characters, using the given condition, find the color.Example Live Demo#include using namespace std; char combination(string s) {    char color = s[0];    for (int i = 1; i < s.length(); i++) {       ... Read More

Suppose we have a number A, and LCM and GCD values, we have to find another number B. If A = 5, LCM is 25, HCF = 4, then another number will be 4. We know that −$$𝐴∗𝐵=𝐿𝐶𝑀∗𝐻𝐶𝐹$$$$𝐵= \frac{LCM*HCF}{A}$$Example Live Demo#include using namespace std; int anotherNumber(int A, int LCM, int GCD) {    return (LCM * GCD) / A; } int main() {    int A = 5, LCM = 25, GCD = 4;    cout

Suppose one array A is given. We have to find a number of surpasser of each element in that array. The surpassers are greater elements which are present at the right side of the array of the current element. Suppose A = {2, 7, 5, 3, 0, 8, 1}, the surpassers are {4, 1, 1, 1, 2, 0, 0}, so 2 has 4 numbers at right side, which are greater than 4, and the same rule for others. The solution is very simple, two nested loops will be there, for each element, it will count surpassers, then store them into ... Read More

Here we will see a program, that can find nth term of the Dragon Curve sequence. The Dragon curve sequence is an infinite binary sequence. It starts with 1, and in each step, it alternatively adds 1s and 0s before and after each element of the previous term, to form the next term.Term 1 : 1Term 2 : 110Term 3 : 1101100Term 4 : 110110011100100We will start with 1, then add 1 and 0, alternatively after each element of the preceding term. When the new term obtained becomes the current term, then repeat the steps from 1 to n to ... Read More

Here we will see how to find the LCM of Rational numbers. We have a list of rational numbers. Suppose the list is like {2/7, 3/14, 5/3}, then the LCM will be 30/1.To solve this problem, we have to calculate LCM of all numerators, then gcd of all denominators, then the LCM of rational numbers, will be like −$$LCM =\frac{LCM\:of\:all\:𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟𝑠}{GCD\:of\:all\:𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟𝑠}$$Example Live Demo#include #include #include using namespace std; int LCM(int a, int b) {    return (a * b) / (__gcd(a, b)); } int numeratorLCM(vector vect) {    int result = vect[0].first;    for (int i = 1; i ... Read More