Career Pivot into Development Journal, Day 6: JS Practice 5 – sumAll.js

takdevelops

TakDevelops

Posted on March 31, 2022

Career Pivot into Development Journal, Day 6: JS Practice 5 – sumAll.js

It's been a few days, I've been quite busy with work so I had to prioritize that first, but I'm so happy I can get back to programming again! :D Moving on to the next exercise...


Understanding the problem

In this exercise, I have to create a function that takes two numbers as parameters, and sums all of the numbers between those two numbers.

Plan

  • Does the program have a UI? What will it look like? What functionality will it have? No UI
  • What are the inputs? Will the user enter in data or will you get it from somewhere else? Two numbers in the function parameter. The numbers will be entered when calling the function
  • What are the outputs? A number, which is the sum of all incremental whole numbers between the two parameters

Pseudocode

Declare a function that takes two parameters, `startNum` and `endNum`. Name it `sumAll`

Create a loop where the initialiser = `startNum` and the condition = `endNum`. Increment the loop.

Declare a variable `sum` and set it equal to 0

Concatenate `i` with `i` and store it in the variable `sum`

Return `sum`

Call the function `sumAll(startNum, endNum)`
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Divide & Conquer

Declare a function that takes two parameters, startNum and endNum. Name it sumAll

const sumAll = function(startNum, endNum) {}
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Create a loop where the initialiser = startNum and the condition = endNum. Increment the loop.

const sumAll = function(startNum, endNum) {
    for (let i = startNum; i <= endNum; i++) {
}
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Declare a variable sum and set it equal to 0

const sumAll = function(startNum, endNum) {
    let sum = 0;
    for (let i = startNum; i <= endNum; i++) {}
}
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Concatenate i with i and store it in the variable sum

const sumAll = function(startNum, endNum) {
    let sum = 0;
    for (let i = startNum; i <= endNum; i++) {
        sum += i;
}
}
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Return sum

const sumAll = function(startNum, endNum) {
    let sum = 0;
    for (let i = startNum; i <= endNum; i++) {
        sum += i;
    }
    return sum;
};
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Call the function sumAll(startNum, endNum)

const sumAll = function(startNum, endNum) {
    let sum = 0;
    for (let i = startNum; i <= endNum; i++) {
        sum += i;
    }
    return sum;
};

sumAll(1,5);
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Putting it to the test

There are 6 tests that I need to pass:

  • sums numbers within the range (1 ms)
  • Works with large numbers
  • Works with larger number first (1 ms)
  • Returns ERROR with negative numbers
  • Returns ERROR with non-number parameters
  • Returns ERROR with non-number parameters (1 ms)

Let's tackle these, starting from all the tests that should return ERROR. The first two tests have already passed based on our code above. To pass the rest of the tests, we need to use if...else statements in our sumAll function

Returns ERROR with negative numbers

if (startNum < 0 || endNum < 0) {
        return 'ERROR';
    }
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Returns ERROR with non-number parameters

if (startNum < 0 || endNum < 0 || typeof(startNum) !== 'number' || typeof(endNum) !== 'number') {
        return 'ERROR';
    }
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Works with larger number first (1 ms)

Working with larger number first means startNum > endNum. For this, we need to reverse the loop that we had above. Instead of incrementing, we will decrement

if (startNum < 0 || endNum < 0 || typeof(startNum) !== 'number' || typeof(endNum) !== 'number') {
        return 'ERROR';
    } else if (startNum > endNum) {
        for (let i = startNum; i >= endNum; i--) {
            sum += i;
        }
        return sum
    }
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Finally, we add in the incremental for loop we had above

const sumAll = function(startNum, endNum) {
    let sum = 0;
    if (startNum < 0 || endNum < 0 || typeof(startNum) !== 'number' || typeof(endNum) !== 'number') {
        return 'ERROR';
    } else if (startNum < endNum) {
        for (let i = startNum; i <= endNum; i++) {
            sum += i;
        }
        return sum;
    } else if (startNum > endNum) {
        for (let i = startNum; i >= endNum; i--) {
            sum += i;
        }
        return sum
    }
};
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✓ sums numbers within the range (1 ms)
✓ works with large numbers
✓ works with larger number first
✓ returns ERROR with negative numbers
✓ returns ERROR with non-number parameters
✓ returns ERROR with non-number parameters

All test have passed! On to the next exercise...

💖 💪 🙅 🚩
takdevelops
TakDevelops

Posted on March 31, 2022

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