Server Side Programming Articles - Page 2164 of 2646

Here we will see how to check whether a number is Mystery number or not. A Mystery number is a number that can be represented by sum of two numbers, and the numbers re reverse of each other. Let us see the code to get better idea. We have to check all pairs and find the decision.Example Live Demo#include using namespace std; int revNum(int str) {    string s = to_string(str);    reverse(s.begin(), s.end());    stringstream ss(s);    int rev = 0;    ss >> rev;    return rev; } bool isMysteryNumber(int n) {    for (int i=1; i

Here we will see, if a number has adjacent set bits in its binary representation. Suppose the number 12 has two consecutive 1s (12 = 1100).To check this type of number, the idea is very simple. We will shift the number 1 bit, then do bitwise AND. If the bitwise AND result is non-zero, then there must be some consecutive 1s.Example Live Demo#include using namespace std; bool hasConsecutiveOnes(int n) {    if((n & (n >> 1)) == 1){    return true;    }else{       return false;    } } int main() {    int num = 67; //1000011    if(hasConsecutiveOnes(num)){       cout

In this section we will check whether a number has same number of set bits and unset bits or not. Suppose the number 12 is there. Its binary representation is 1100. This has same number of 0s and 1s.The approach is simple. We will check each bit of the number, and if that is 1, then increase set_bit_count, and if it is 0, then increase unset_bit_count. Finally, if they are same, then return true, otherwise false.Example Live Demo#include using namespace std; bool hasSameSetUnset(int n) {    int set_count = 0, unset_count = 0;    while(n){       if((n & ... Read More

In this section we will see, whether a number can be represented as tree consecutive numbers or not. Suppose a number is 27. This can be represented as 8 + 9 + 10.This can be solved in two different approach. The first approach is Naïve approach. In that approach, we have to check i + (i + 1) + (i + 2) is same as number or not. Another efficient approach is by checking whether the number is divisible by 3 or not. Suppose a number x can be represented by three consecutive 1s, then x = (y - 1) ... Read More

In this article, our task is to check if we can represent a given number as the sum of two non-zero powers of 2 in C++. For this, we will check whether the given number N can be represented as (2^x + 2^y) where x, y > 0. Here are some example scenarios: Scenario 1 Input:10 Output: Can be represented Explanation: 10 = 8 + 2 10 = 2^3 + 2^1 => 10 can be represented as 2x + 2y Scenario 2 Input:3 Output: Can be represented Explanation: 10 = 2 + ... Read More

In this section, we will see if we can express one number as the sum of two triangular numbers or not. The triangular numbers are like below −From the example, we can see that 1, 3, 6, 10 are some triangular numbers. We need to express a number N (say 16) as sum of two triangular numbers (6, 10).The approach is very simple. We have to get all triangular numbers less than N. Form a set from these values. Now we have to take a number say X from the set, and check whether N – X is present in ... Read More

Here we will check whether we can represent a number as power, like xy or not. Suppose a number 125 is present. This can be represented as 53. Another number 91 cannot be represented as power of some integer value.AlgorithmisRepresentPower(num): Begin    if num = 1, then return true    for i := 2, i2

Here we will check whether we can represent a number as power, like ab or not. Suppose a number 125 is present. This can be represented as 53. Another number 91 cannot be represented as power of some integer value.AlgorithmisRepresentPower(num): Begin    if num = 1, then return true    for i := 2, i2

Here we will check whether we can represent a number like ab or not. Suppose a number 125 is present. This can be represented as 53. Another number 91 cannot be represented as power of some integer value.AlgorithmisRepresentPower(num): Begin    if num = 1, then return true    for i := 2, i2

Here we will see if one number can be represented as sum of two or more consecutive numbers or not. Suppose a number is 12. This can be represented as 3+4+5.There is a direct and easiest method to solve this problem. If a number is power of 2, then it cannot be expressed as sum of some consecutive numbers. There are two facts that we have to keep in mind.Sum of any two consecutive numbers is odd, then one of them will be odd, another one is even.Second fact is 2n = 2(n-1) + 2(n-1).Example Live Demo#include using namespace std; ... Read More