Server Side Programming Articles - Page 2166 of 2646

Here we will see how to check a number is divisible by 25 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 25, when the last two digits are 00, or they are divisible by 25.Example Live Demo#include using namespace std; bool isDiv25(string num){    int n = num.length();    int last_two_digit_val = (num[n-2] - '0') * 10 + ((num[n-1] - '0'));    if(last_two_digit_val % 25 == 0)       return true;       return false; } int main() {    string num = "451851549333150";    if(isDiv25(num)){       cout

Here we will see how to check a number is divisible by 20 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 20, when that is divisible by 10, and after dividing 10, the remaining number is divisible by 2. So the case is simple. If the last digit is 0, then it is divisible by 10, and when it is divisible by 10, then the second last element is divisible by 2, the number is divisible by 20.Example Live Demo#include using namespace std; bool isDiv20(string ... Read More

Here we will see how to check a number is divisible by 2, 3 and 5 or not. In this case the number is very large number. So we put the number as string.A number will be divisible by 2, 3 and 5 if that number is divisible by LCM of 2, 3 and 5. So the LCM of 2, 3, 5 is 30. We have to check the number is divisible by 30 or not. A number is divisible by 30 when it is divisible by 10 (last digit is 0) and divisible by 3 (sum of all digits ... Read More

In this article, we are given a larger number and we need to check whether it is divisible by 11 or not using the C++ program. To handle large numbers, we treat the number as a string and apply a divisibility rule to check if it's divisible by 11. Rule of Divisibility by 11 A number is divisible by 11 if the difference between the sum of its digits at odd positions and the sum of its digits at even positions is divisible by 11. Consider the following example scenarios to understand the divisibility of a large number ... Read More

Here we will see how to check a number is divisible by 15 or not. In this case the number is very large number. So we put the number as string.To check whether a number is divisible by 15, if the number is divisible by 5, and divisible by 3. So to check divisibility by 5, we have to see the last number is 0 or 5. To check divisibility by 3, we will see the sum of digits are divisible by 3 or not.Example Live Demo#include using namespace std; bool isDiv15(string num){    int n = num.length();    if(num[n ... Read More

Here we will see how to check a string is made up of alternating characters or not. If a string is like XYXYXY, it is valid, if a string is like ABCD, that is invalid.The approach is simple. We will check whether all ith character and i+2 th character are same or not. if they are not same, then return false, otherwise return true.Example Live Demo#include using namespace std; bool hasAlternateChars(string str){    for (int i = 0; i < str.length() - 2; i++) {       if (str[i] != str[i + 2]) {          return ... Read More

In this article, we will learn about the solution and approach to solve the given problem statement.Problem statementGiven a range, we need to print all the even numbers in the given range.The brute-force approach is discussed below −Here we apply a range-based for loop which provides all the integers available in the input interval.After this, a check condition for even numbers is applied to filter all the odd numbers.This approach takes O(n) + constant time of comparison.Now let’s see the implementation below −Examplestart, end = 10, 29 # iteration for num in range(start, end + 1):    # check   ... Read More

In this article, we will learn about the solution and approach to solve the given problem statement.Problem statementGiven a number n, we need to check whether the given number is a power of two.ApproachContinue dividing the input number by two, i.e, = n/2 iteratively.We will check In each iteration, if n%2 becomes non-zero and n is not 1 then n is not a power of 2.If n becomes 1 then it is a power of 2.Let’s see the implementation below −Exampledef isPowerOfTwo(n):    if (n == 0):       return False    while (n != 1):       ... Read More

In this article, we will learn about the solution and approach to solve the given problem statement.Problem statementGiven an input string we need to find the most occurring character and its count.ApproachCreate a dictionary using Counter method having strings as keys and their frequencies as values.Find the maximum occurrence of a character i.e. value and get the index of it.Now let’s see the implementation below −Examplefrom collections import Counter    def find(input_):    # dictionary    wc = Counter(input_)    # Finding maximum occurrence    s = max(wc.values())    i = wc.values().index(s)    print (wc.items()[i]) # Driver program if __name__ ... Read More

In this article, we will learn about the solution and approach to solve the given problem statement.Problem statementGiven a dictionary, we need to find the three highest valued values and display them.Approach 1 − Using the collections module ( Counter function )Example Live Demofrom collections import Counter # Initial Dictionary my_dict = {'t': 3, 'u': 4, 't': 6, 'o': 5, 'r': 21} k = Counter(my_dict) # Finding 3 highest values high = k.most_common(3) print("Dictionary with 3 highest values:") print("Keys: Values") for i in high:    print(i[0], " :", i[1], " ")OutputDictionary with 3 highest values: Keys: Values r : 21 t ... Read More