• Feb 4, 2021 •aedrarian
0 likes • 0 views
#include <iostream> using namespace std; main { cout << "No tabbing. That's very sad :(\n"; cout << "No in-editor highlighting either :(((\n"; cout << "Descriptions might be niice too."; }
• Aug 31, 2020 •joshwrou
1 like • 3 views
#include <iostream> using namespace std; int main() { cout << "Hello World!\n"; // Prints out "Hello World" return 0; }
• Sep 7, 2022 •LeifMessinger
#include <iostream> #include <cstring> int main(int argc, char** argv){ //With decimal if(strstr(argv[1], ".") != nullptr){ int i = 0; //Skip i to first non 0 digit while(argv[1][i] < '1' || argv[1][i] > '9') ++i; //If digit comes before decimal if((argv[1] + i) < strstr(argv[1], ".")){ //Good example of pointer arithmetic std::cout << strlen(argv[1] + i) - 1 << std::endl; //Another good example }else{ //If digit is after decimal std::cout << strlen(argv[1] + i) << std::endl; } }else{ //Without decimal int m = 0; int i = 0; while(argv[1][i] < '1' || argv[1][i] > '9') ++i; //In case of some number like 0045 for(; argv[1][i] != '\0'; ++i){ if(argv[1][i] >= '1' && argv[1][i] <= '9') m = i + 1; } std::cout << m << std::endl; } return 0; }
• Dec 24, 2021 •aedrarian
3 likes • 21 views
/* Good morning! Here's your coding interview problem for today. This problem was asked by Stripe. Given an array of integers, find the first missing positive integer in linear time and constant space. In other words, find the lowest positive integer that does not exist in the array. The array can contain duplicates and negative numbers as well. For example, the input [3, 4, -1, 1] should give 2. The input [1, 2, 0] should give 3. You can modify the input array in-place. */ #include <iostream> using namespace std; int calcMissing(int* input, int size) { int sum = 0; int n = 1; //add one to account for missing value for(int i = 0; i < size; i++) { if(input[i] > 0) { sum += input[i]; n++; } } //If no numbers higher than 0, answer is 1 if(sum == 0) return 1; return (n*(n+1)/2) - sum; //Formula is expectedSum - actualSum /* expectedSum = n*(n+1)/2, the formula for sum(1, n) */ } int main() { cout << calcMissing(new int[4]{3, 4, -1, 1}, 4) << endl; cout << calcMissing(new int[3]{1, 2, 0}, 3) << endl; //No positive numbers cout << calcMissing(new int[1]{0}, 1) << endl; }
• Aug 5, 2023 •usama
1 like • 6 views
• Nov 18, 2022 •AustinLeath
0 likes • 1 view
#include <iostream> #include <cmath> #include <string.h> using namespace std; int main() { string tickerName; int numOfContracts; float currentOptionValue; cout << "Enter a stock ticker: "; getline(cin, tickerName); cout << "Enter the current number of " << tickerName << " contracts you are holding: "; cin >> numOfContracts; cout << "Enter the current price of the option: "; cin >> currentOptionValue; cout << "The value of your " << tickerName << " options are: $" << (currentOptionValue * 100.00) * (numOfContracts); cout << endl; return 0; }