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Server Side Programming Articles
Page 1448 of 2109
A Product Array Puzzle in C++?
Here we will see one interesting problem related to array. There is an array with n elements. We have to create another array of n elements. But the i-th position of second array will hold the product of all elements of the first array except the i-th element. And one constraint is that we cannot use the division operator in this problem.If we can use the division, operation, we can easily solve this problem, by getting the product of all elements, then divide i-th element of first array and store it into i-th place of the second array.Here we are ...
Read MoreAbsolute Difference of even and odd indexed elements in an Array (C++)?
Here we will see how we can get the absolute differences of odd and even indexed elements in an array. The absolute difference indicates that if the difference of one pair is negative, the absolute value will be taken. For an example, let the numbers are {1, 2, 3, 4, 5, 6, 7, 8, 9}. So the even position elements are 1, 3, 5, 7, 9 (starting from 0), and odd place elements are 2, 4, 6, 8. So the difference for even placed data are |1 - 3| = 2, then |2 - 5| = 3, |3 - 7| ...
Read MoreC++ Program for Common Divisors of Two Numbers?
Here we will see how we can get the number of common divisors of two numbers. We are not going to find all common divisors, but we will count how many common divisors are there. If two numbers are like 12 and 24, then common divisors are 1, 2, 3, 4, 6, 12. So there are 6 common divisors, so the answer will be 6.AlgorithmcountCommonDivisor(a, b)begin count := 0 gcd := gcd of a and b for i := 1 to square root of gcd, do if gcd is divisible by 0, then ...
Read MoreC++ Program for GCD 0.of more than two (or array) numbers?
Here we will see how we can get the gcd of more than two numbers. Finding gcd of two numbers are easy. When we want to find gcd of more than two numbers, we have to follow the associativity rule of gcd. For example, if we want to find gcd of {w, x, y, z}, then it will be {gcd(w, x), y, z}, then {gcd(gcd(w, x), y), z}, and finally {gcd(gcd(gcd(w, x), y), z)}. Using array it can be done very easily.Algorithmgcd(a, b)begin if a is 0, then return b end if return gcd(b ...
Read MoreC++ Program for Smallest K digit number divisible by X?
In this problem we will try to find smallest K-digit number, that will be divisible by X. To do this task we will take the smallest K digit number by this formula (10^(k-1)). Then check whether the number is divisible by X or not, if not, we will get the exact number by using this formula.(min+ 𝑋)−((min+ 𝑋) 𝑚𝑜𝑑 𝑋)One example is like a 5-digit number, that is divisible by 29. So the smallest 5-digit number is 10000. This is not divisible by 29. Now by applying the formula we will get −(10000+ 29)−((10000+29) 𝑚𝑜𝑑 29)=10029−24=10005The number 10005 is divisible ...
Read MoreC++ program to concatenate a string given number of times?
Here we will see how we can concatenate a string n number of times. The value of n is given by the user. This problem is very simple. In C++ we can use + operator for concatenation. Please go through the code to get the idea.AlgorithmconcatStrNTimes(str, n)begin res := empty string for i in range 1 to n, do res := concatenate res and res done return res endExample#include using namespace std; main() { string myStr, res = ""; int n; cout > myStr; cout > n; for(int i= 0; i < n; i++) { res += myStr; } cout
Read MoreC++ Program to count Vowels in a string using Pointer?
To get the vowels from a string, we have to iterate through each character of the string. Here we have to use pointers to move through the string. For this we need C style strings. If the string is pointed by str, then *str will hold the first character at the beginning. Then if str is increased, the *str will point next character and so on. If the character is in [a, e, i, o, u] or [A, E, I, O, U] then it is vowel. So we will increase the countAlgorithmcountVowels(str)begin count := 0 for each character ...
Read MoreC++ Program to find the sum of a Series 1/1! + 2/2! + 3/3! + 4/4! + ...... n/n!
Here we will see how we can get the sum of the given series. The value of n will be given by user. We can solve this problem by making a factorial function, and get factorial in each step in the loop. But factorial calculation is costlier task than normal addition. We will use the previous factorial term in the next one. Like 3! is (3 * 2 * 1), and 4! is 4 * 3!. So if we store 3! into some variable, we can use that and add the next number only to get the next factorial easily.Algorithmsum_series_fact(n)begin ...
Read MoreC++ Program for Gnome Sort?
Here we will see how the gnome sort works. This is another sorting algorithm. In this approach if the list is already sorted it will take O(n) time. So best case time complexity is O(n). But average case and worst case complexity is O(n^2). Now let us see the algorithm to get the idea about this sorting technique.AlgorithmgnomeSort(arr, n)begin index := 0 while index < n, do if index is 0, then index := index + 1 end if if arr[index] >= arr[index -1], then ...
Read MoreC++ Program for Largest K digit number divisible by X?
In this problem we will try to find largest K-digit number, that will be divisible by X. To do this task we will take the largest K digit number by this formula ((10^k) – 1). Then check whether the number is divisible by X or not, if not, we will get the exact number by using this formula.𝑚𝑎𝑥−(𝑚𝑎𝑥 𝑚𝑜𝑑 𝑋)One example is like a 5-digit number, that is divisible by 29. So the largest 5-digit number is 99999. This is not divisible by 29. Now by applying the formula we will get −99999−(99999 𝑚𝑜𝑑 29)=99999−7=99992The number 99992 is divisible by ...
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