Server Side Programming Articles - Page 1941 of 2650

This article aims to print a pyramid pattern by using the recursive implementation of C++ programming. Here is the algorithm as following to do so;AlgorithmStep-1 Set the height of the pyramid Step-2 Adjust space using recursion function Step-3 Adjust Hash(#) character using recursion function Step-4 Call both functions altogether to print the Pyramid patternExampleAs said the above algorithm, the following genuine C++ code economics is written as following; Live Demo#include using namespace std; // function to print spaces void print_space(int space){    if (space == 0)       return;    cout

This article demonstrate the functioning of proj() in order to perform projection upon complex numbers. Here the syntax of the proj() method in c++ programming as follows;template complex proj (const complex& z);ExampleThe proj() method takes a parameter as argument which represent the complex number and returns the projection of complex number described below in the sample as; Live Demo#include #include using namespace std; int main(){    std::complex c1(3, 5);    cout

Suppose we have some rocks, each rock has a positive integer weight. In each turn, we will take two heaviest rocks and smash them together. consider the stones have weights x and y and x 1:          stones.sort()          s1,s2=stones[-1],stones[-2]          if s1==s2:             stones.pop(-1)             stones.pop(-1)          else:             s1 = abs(s1-s2)             stones.pop(-1)             stones[-1] = s1       if len(stones):          return stones[-1]       return 0 ob1 = Solution() print(ob1.lastStoneWeight([2,7,4,1,6,1]))Input[2,7,4,1,6,1]Output1

Suppose there are 2N persons. A company wants to organize one interview. The cost for flying the i-th person to city A is costs[i][0], and the cost for flying the i-th person to city B is costs[i][1]. We have to find the minimum cost to fly every person to a city, such that N people arrive in every city. So if the given list is [[10, 20], [30, 200], [400, 50], [30, 20]] The output will be 110. So we will send the person P1 to city A with cost 10, Second person to city A with cost 30, third ... Read More

Suppose we have an array A of integers, our output will be true if and only if we can partition the array into three non-empty parts whose sum is equal.Formally, we can partition the array if we can find the indexes i+1 < j with (A[0] + A[1] + ... + A[i] is same as A[i+1] + A[i+2] + ... + A[j-1] and A[j] + A[j-1] + ... + A[A.length - 1])So if the input is [0, 2, 1, -6, 6, -7, 9, 1, 2, 0, 1], then the output will be true. Three arrays will be [0, 2, 1], ... Read More

Suppose we have a list of songs, the i-th song has a duration of time[i] seconds. We have to find the number of pairs of songs for which their total time in seconds is divisible by 60.So if the time array is like [30, 20, 150, 100, 40], then the answer will be 3. Three pairs will be (3, 150), (20, 100), (20, 40) for all cases the total duration is divisible by 60.To solve this, we will follow these steps −Take a map rem to store remainders. Set ans := 0for all elements i in time −if i is ... Read More

Suppose we have an array of integers called A, and an array queries. For the i-th query value = queries[i][0] and index = queries[i][1], we will add value to A[index]. Then, the answer of the i-th query is the sum of the even values of A. We have to find the answer to all queries. We will find an array, that should have answer[i] as the answer to the i-th query. So if the array is like [1, 2, 3, 4], and the query array is like [[1, 0], [-3, 1], [-4, 0], [2, 3]], then the answer array will ... Read More

Suppose a man is typing some name on keyboard. Sometimes some buttons are long-pressed by mistake. So it may type one or more extra character. So we will take two strings, and check whether the second string is long-pressed name or not. So if the name is “Amit”, and second string is “Ammittt” is longpressed name. But “Ammttt” is not, because character i is not present here.To solve this, we will follow these steps −let j := 0for i := 0, i < second.size, increase i by 1 −if j < actual_name.size and actual_name[j] = second[i], thenincrease j by 1return ... Read More

Suppose we have a string with letters and numbers. We have to generate all possible combinations of that string by taking uppercase and lowercase versions of letters that are present in the string. So if one string has only numbers, only that will be returned. Suppose the string is like “1ab2”, then the strings will be [“1ab2”, “1Ab2”, “1aB2”, “1AB2”]To solve this problem, we will use recursive approach. It takes the index parameter to start work from that index. It also takes a temp string up to which the result is created. When the index is same as the string ... Read More

Suppose we have a string J that indicates some letters that are considered as Jewel, and another string S, that represents some stones that we have. Our task is to find how many of stones in S is also jewel. The letters in J and S are case sensitive. So if the J = “aZc”, and S = “catTableZebraPicnic” then there are 7 jewels.To solve this we will convert the string into a list of characters. If the character in J present in S, then increase the count.ExampleLet us see the following implementation to get better understanding − Live Democlass Solution(object): ... Read More