# Backend Interview questions

## Exercise 1

Given an integer called `sum` and an array of integers called `numbers`, check whether there’s a combination of any two elements in the `numbers` array that can be added to equal `sum` exactly.

### Example:

* **Input:** `sum = 17`, `numbers = [1, 9, 2, 8]`, **Answer:** `true` since `9 + 8 = 17`
* **Input:** `sum = 4`, `numbers = [1, 9, 2, 8]`, **Answer:** `false`

<!-- ### Answer:

function checkSumInArray(sum, numbers) {
    const parsedNumbers = {};
    for (let i = 0; i < numbers.length; i++) {
        const diff = sum - numbers[i];
        if (parsedNumbers[diff]) {
            return true;
        } else {
            parsedNumbers[numbers[i]] = true;
        }
    };
    return false;
}

## Exercise 2

Your company assigns each customer a membership ID, and you are implementing a check digit for those IDs. The check digit should be calculated by adding up all digits in each membership ID. If the result of the sum is a number with more than a single digit, another iteration is required, and the digits of the result also should be added together. This process should repeat until a single-digit number is calculated.

### Example:

For the membership ID `"55555"` the sum of all digits is `25`. Because this is not a single-digit number, `2` and `5` would be added, and the result, `"7"`, would be the check digit.

<!-- ### Answer:

function checkDigit(num) {
    let tot = 0;
    let splity = num.split('');

	  for(let i = 0; i < splity.length; i++) {
        tot += parseInt(splity[i]);
    }

    if (tot > 9) {
        return checkDigit(tot.toString());
    } else {
        return num;
    }
}

-->
## Exercise 3

You are climbing stairs. It takes `n` steps to get to the top. Each time, you can take 1 step or 2 steps. How many distinct ways to get to the top?

### Example:
  

* **Input:** `n = 1`, **Answer:** `1` | **Input:** `n = 2`, **Answer:** `2` | **Input:** `n = 3`, **Answer:** `3`

<!-- ### Answer:

function climbStairs(n) {
    if (n == 1) return 1;
    if (n == 2) return 2;
    return climbStairs(n - 1) + climbStairs(n - 2);
}