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Math.clz32()The Math.clz32() function returns the number of leading zero bits in the 32-bit binary representation of a number. In a binary representation, there are 32 numbers consisting of only 1's and 0's. This method scrutinizes through each and every element and returns the number of 0's.syntaxMath.clz32(number);Example-1In the following example, numbers 1 and 0 were passed into the Math.Clz32() function and the number of trailing zeroes were displayed in the output.Live Demo document.write(Math.clz32(1)); document.write(""); document.write(Math.clz32(0)); Output31 32Example-2In the following example, numbers 21 and 20 were passed into the ... Read More
Here we will see some amazing results by using the system() function in C or C++. The system function is present in Windows, Linux and MAC operating systems. This function is used to execute the system commands that can be written in Command line.Here we will see two usages if system function in C or C++. The first one is getting the IP configuration details using C++ program.Example#include #include using namespace std; int main() { system("C:\Windows\System32\ipconfig"); }OutputWindows IP Configuration Ethernet adapter Local Area Connection: Connection-specific DNS Suffix . : domain.name Link-local IPv6 Address . . ... Read More
In this section we will see how to generate all possible strings of any length, this will take each combination of characters to make string. For example, if the string is ABC, then it will generate − {A, B, C, AB, BA, BC, CB, CA, AC, ABC, ACB, BAC, BCA, CAB, CBA}Let us see the example to get the idea.AlgorithmprintAllString(str)Begin n := length of the string str count is 2^n – 1 for each number 0 to count, do sub_str := empty string for j in range 0 to n, do ... Read More
Here we will see one problem, We have N and base B. Our task is to count all N digit numbers of base B without any leading 0s. So if N is 2 and B is 2 there will be four numbers 00, 01, 10, 11. So only two of them are valid for this section. These are 10, 11, there are no leading 0s.If the base is B, then there are 0 to B – 1 different digits. So BN number of different N digit values can be generated (including leading 0s). The first digit is 0m if we ... Read More
Here we will see how to count number of co-prime pairs from the range, where a number will not appear more than a single pair.Before discussing the logic, let us see what are the co-prime numbers? The co-prime numbers are those numbers which has only one positive integer divisor, that is 1. In other words, we can say the GCD of these two numbers is 1.Here we are providing the lower and upper limit. If the lower and upper limits are 1 and 6, then there are three pairs. These are (1, 2), (3, 4) and (5, 6)The approach for ... Read More
Here we will see all possible binary numbers of n bit (n is given by the user) where the sum of each half is same. For example, if the number is 10001 here 10 and 01 are same because their sum is same, and they are in the different halves. Here we will generate all numbers of that type.AlgorithmgenAllBinEqualSumHalf(n, left, right, diff)left and right are initially empty, diff is holding difference between left and rightBegin if n is 0, then if diff is 0, then print left + right ... Read More
Here we will see one simple problem. We have to find all numbers that are palindrome in nature in a given list. The approach is simple, take each number from list and check it is palindrome or not, and print the number.AlgorithmgetAllPalindrome(arr, n)Begin for each element e in arr, do if e is palindrome, then print e end if done EndExample#include #include using namespace std; bool isPalindrome(int n){ int reverse = 0, t; t = n; while (t != 0){ ... Read More
We're given two singly linked lists, where each node stores one digit of a number. The digits are arranged from left to right, just like how we normally write numbers. For example: 7 -> 2 -> 4 -> 3 represents the number 7243. Our task is to add these two numbers and return the sum as a new linked list in the same (left-to-right) order. The input number can contain zeroes at the start, but in the output, there should not be any leading zeros. Let's understand this with a diagram given below - Scenario 1 Input: List 1 ... Read More
Here we will see how to calculate the number of trailing 0s for the result of factorial of any number. So if the n = 5, then 5! = 120. There is only one trailing 0. For 20!, it will be 4 zeros as 20! = 2432902008176640000.The easiest approach is just calculating the factorial and count the 0s. But this approach fails for large value of n. So we will follow another approach. The trailing zeros will be there, if the prime factors are 2 and 5. If we count the 2s and 5s we can get the result. For ... Read More
In this section we will see if one array is given with n numbers, we have to check if we make a number using all of the elements of these numbers, that number will be divisible by 3 or not. If the array elements are {15, 24, 23, 13}, then the we can make integer like 15242313. It will be divisible by 3.AlgorithmcheckDivThree(arr)Begin rem := 0 for each element e in arr, do rem := (rem + e) mod 3 done if rem is 0, then return true end ... Read More