Java

Best Way to Check if a Number Contains Another Number 2022-06-19

Need to find out if a number contains another number? And you want better performance than converting to a string and then looping through that to find the key number? Well you’ve come to the right place!

Using a combination of the modulo operator and division it’s possible to pull the number apart one digit at a time until it reaches 0.

Implementation Using a Loop

def checkNumberIfContainsKey(number, key):
    while number > 0:
        if number % 10 == key:
             return True
        number = number // 10
    return False

package main

import "fmt"

func checkNumberIfContainsKey(number, key int) bool {
	for number > 0 {
		if number%10 == key {
			return true
		}
		number = number / 10
	}
	return false
}

func main() {
	fmt.Println(checkNumberIfContainsKey(359, 9))
}

function checkNumberIfContainsKey(number, key){
    while(number > 0){
        if(number%10 == key){
            return true;
        }
        number = Math.floor(number / 10);        
    }
    return false;
}

#include "bits/stdc++.h"
using namespace std;

bool checkNumberIfContainsKey(int number, int key)
{
	while (number > 0) 
	{
		if (number % 10 == key) 
		{
			return true;
		}
		number = number / 10;
	}
	return false;
}

int main(int argc, char const *argv[])
{
	cout << checkNumberIfContainsKey(359,9) << endl;
	return 0;
}

class ContainsKey {
    public static boolean checkNumberIfContainsKey(int num, int key) {
        while (num > 0) {
            if (num % 10 == key) {
                return true;
            }
            num = num / 10;
        }
        return false;
    }
}

Recursive Implementation

def checkNumberIfContainsKey(number, key):
    if number == 0:
        return False
    
    if number % 10 == key:
        return True
    
    return checkNumberIfContainsKey(number // 10, key)

package main

import "fmt"

func checkNumberIfContainsKey(number, key int) bool {
	if number == 0 {
		return false
	}
	if number%10 == key {
		return true
	}
	return checkNumberIfContainsKey(number / 10, key)
	
}

func main() {
	fmt.Println(checkNumberIfContainsKey(359, 9))
}

function checkNumberIfContainsKey(number, key){
    if (number === 0) {
    	return false;
    }
    if(number%10 == key) {
    	return true;
    }
    return checkNumberIfContainsKey(Math.floor(number / 10), key);
}

#include "bits/stdc++.h"
using namespace std;

bool checkNumberIfContainsKey(int number, int key)
{
	while (number > 0) 
	{
		if (number % 10 == key) 
		{
			return true;
		}
		number = number / 10;
	}
	return false;
}

int main(int argc, char const *argv[])
{
	cout << checkNumberIfContainsKey(359,9) << endl;
	return 0;
}

class ContainsKey {
    public static boolean checkNumberIfContainsKey(int num, int key) {
        if (num == 0) {
            return false;
        }

        if (num % 10 == key) {
            return true;
        }

        return checkNumberIfContainsKey(num / 10, key);
    }
}

Two Sum 2022-06-19
Given an array of unsorted numbers nums and an integer target, find two integers in the array that sum to the target and return their indices.
There are three ways that I know of to solve this problem. Below you’ll find a description of each with some brief code examples. I would like to encourage you to try to implement your own solution first before scrolling down.
Solution 1: Brute Force
The first way, which is the brute force method, is to use nested loops. It tries every possible combination by looping over and take exponential time.

Implementation Using a Loop

Recursive Implementation

Solution 1: Brute Force